Basic Algebra and Logarithms for JEE: Modulus, Inequalities and Logs

Ritesh Raj · 3 Nov 2026

30 min read27 sectionsConcept

Basic algebra is the part of mathematics that every other chapter uses without saying so. An inequality decides a domain, a modulus decides a number of roots, a logarithm turns a power into a product, and one identity can replace a page of expansion. JEE rarely asks this chapter on its own, but these methods sit inside questions from almost every unit, so speed here keeps the rest short.

This note covers numbers and intervals, indices and surds, identities, polynomials, inequalities by the wavy curve method, the modulus, the greatest integer, logarithms and exponential equations. Every worked example is a JEE Main 2025 or 2026 question unless it says otherwise, and every answer was re-derived from scratch.

Numbers and intervals

\mathbb{N}\subset\mathbb{W}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}

A rational number is $\dfrac pq$ with $p,q$ integers and $q\ne0$ ; its decimal ends or recurs. An irrational number has a decimal that neither ends nor recurs: $\sqrt2$ , $\sqrt3$ , $\pi$ , $e$ .

Combination	Result
rational $\pm$ irrational	always irrational
non-zero rational $\times$ irrational	always irrational
irrational $\pm$ irrational	can be anything: $\sqrt2+(-\sqrt2)=0$
irrational $\times$ irrational	can be anything: $\sqrt2\cdot\sqrt8=4$

Intervals

Interval	Meaning
$(a,b)$	$a<x<b$
$[a,b]$	$a\le x\le b$
$[a,b)$	$a\le x<b$
$(a,b]$	$a<x\le b$
$(a,\infty)$	$x>a$
$(-\infty,b]$	$x\le b$

Four number lines one below the other, each with a highlighted segment from a to b: the first with both end points hollow, the second with both end points filled, the third with a filled left end and a hollow right end, the fourth with a hollow left end and a filled right end

A filled dot means the end point belongs to the set; a hollow dot means it does not. The intersection of two intervals runs from the larger left end to the smaller right end. A set of solutions is always reported as a union of intervals in increasing order.

Infinity takes a round bracket

$\infty$ is not a number and is never reached, so $[2,\infty]$ is wrong. Write $[2,\infty)$ .

Example 1 (IITianForum). (a) Which of these are rational?

(3+\sqrt2)-\sqrt2,\quad \sqrt2\cdot\sqrt{18}

(1+\sqrt2)^2,\quad \frac{\sqrt3}{\sqrt{12}}

(b) For $A=[-2,5)$ and $B=(1,8]$ , write $A\cap B$ , $A\cup B$ and $A-B$ .

▸ SHOW SOLUTION

Step 1. Part (a)

(3+\sqrt2)-\sqrt2=3

\sqrt2\cdot\sqrt{18}=\sqrt{36}=6

(1+\sqrt2)^2=3+2\sqrt2

\frac{\sqrt3}{\sqrt{12}}=\sqrt{\tfrac14}=\tfrac12

So the first, second and fourth are rational; $3+2\sqrt2$ is irrational.

Step 2. Part (b)

Larger left end $1$ , smaller right end $5$ :

A\cap B=(1,5),\qquad A\cup B=[-2,8]

$A-B$ keeps the points of $A$ not in $B$ : $A-B=[-2,1]$ . The point $1$ stays, since $1\notin B$ .

Indices and surds

For $a,b>0$ and real $m,n$ :

Law	Statement
product	$a^m\cdot a^n=a^{m+n}$
quotient	$\dfrac{a^m}{a^n}=a^{m-n}$
power	$(a^m)^n=a^{mn}$
same power	$a^m b^m=(ab)^m$
zero, negative	$a^0=1$ , $a^{-n}=\dfrac1{a^n}$
root	$a^{p/q}=\sqrt[q]{a^p}$

For $a>0$ , $a\ne1$ : $a^x=a^y$ gives $x=y$ . To compare two powers, bring them to the same exponent or the same base.

Surds. $\sqrt a\sqrt b=\sqrt{ab}$ holds for $a,b\ge0$ only. To remove a surd from a denominator, multiply by the conjugate:

\frac{1}{\sqrt a+\sqrt b}=\frac{\sqrt a-\sqrt b}{a-b}

Square root of a surd. If $x+y=a$ and $xy=b$ with $x>y>0$ , then

\sqrt{a+2\sqrt b}=\sqrt x+\sqrt y

\sqrt{a-2\sqrt b}=\sqrt x-\sqrt y

Comparing parts. If $p+q\sqrt n=r+s\sqrt n$ with $p,q,r,s$ rational and $\sqrt n$ irrational, then $p=r$ and $q=s$ .

Example 2 (IITianForum). Which is larger? (a) $2^{300}$ or $3^{200}$ ; (b) $\sqrt7-\sqrt6$ or $\sqrt6-\sqrt5$ .

▸ SHOW SOLUTION

Step 1. Same exponent

2^{300}=(2^3)^{100}=8^{100}

3^{200}=(3^2)^{100}=9^{100}

So $3^{200}$ is larger.

Step 2. Use the conjugate

\sqrt7-\sqrt6=\frac{1}{\sqrt7+\sqrt6}

\sqrt6-\sqrt5=\frac{1}{\sqrt6+\sqrt5}

The first denominator is larger, so the first fraction is smaller.

Answer: (a) $3^{200}$ ; (b) $\sqrt6-\sqrt5$

Example 3 (IITianForum). Simplify $\sqrt{7+4\sqrt3}$ and $\sqrt{11-6\sqrt2}$ . Hence find $\sqrt{7+4\sqrt3}+\sqrt{7-4\sqrt3}$ .

▸ SHOW SOLUTION

Step 1. First surd

$7+4\sqrt3=7+2\sqrt{12}$ . Two numbers with sum $7$ and product $12$ : $4$ and $3$ .

\sqrt{7+4\sqrt3}=\sqrt4+\sqrt3=2+\sqrt3

Step 2. Second surd

$11-6\sqrt2=11-2\sqrt{18}$ . Sum $11$ , product $18$ : $9$ and $2$ .

\sqrt{11-6\sqrt2}=3-\sqrt2

The larger number comes first, so the result is positive.

Step 3. The sum

\sqrt{7-4\sqrt3}=2-\sqrt3

(2+\sqrt3)+(2-\sqrt3)=4

Answer: $2+\sqrt3$ , $3-\sqrt2$ , and the sum is $4$

Example 4 (IITianForum). Find $x<-1$ such that

\left(1+\frac1x\right)^{x+1}=\left(1+\frac1{50}\right)^{50}

▸ SHOW SOLUTION

Step 1. Put x = −y

With $y>1$ :

1+\frac1x=\frac{y-1}{y},\qquad x+1=-(y-1)

Step 2. Rewrite the left side

\left(\frac{y-1}{y}\right)^{-(y-1)}=\left(\frac{y}{y-1}\right)^{y-1}

=\left(1+\frac{1}{y-1}\right)^{y-1}

Step 3. Compare

Matching the two sides, $y-1=50$ , so $y=51$ .

Answer: $x=-51$

Example 5 (JEE Main 2025, 29 January Shift 1). The number of solutions of the equation

\left(\frac9x-\frac9{\sqrt x}+2\right)\left(\frac2x-\frac7{\sqrt x}+3\right)=0

Tap an option to check your answer

(1)

2

(2)

4

(3)

1

(4)

3

▸ SHOW SOLUTION

Step 1. Substitute

Put $t=\dfrac1{\sqrt x}$ , so $t>0$ and $\dfrac1x=t^2$ :

(9t^2-9t+2)(2t^2-7t+3)=0

(3t-1)(3t-2)(2t-1)(t-3)=0

Step 2. Back to x

$t=\tfrac13,\tfrac23,\tfrac12,3$ , all positive. Then $x=\dfrac1{t^2}$ :

x=9,\ \tfrac94,\ 4,\ \tfrac19

Four distinct values.

Answer: option (2)

Identities and factorisation

(a+b)^3=a^3+b^3+3ab(a+b)

a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)

(a+b+c)^2=\textstyle\sum a^2+2\sum ab

a^4+a^2b^2+b^4

=(a^2+ab+b^2)(a^2-ab+b^2)

The most used identity of the chapter:

a^3+b^3+c^3-3abc=(a+b+c)\,Q

Q=a^2+b^2+c^2-ab-bc-ca

Q=\tfrac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]

So $Q\ge0$ , and $Q=0$ only when $a=b=c$ .

Condition	Consequence
$a+b+c=0$	$a^3+b^3+c^3=3abc$
$a^3+b^3+c^3=3abc$	$a+b+c=0$ or $a=b=c$
$a^2+b^2+c^2=ab+bc+ca$	$a=b=c$

Ratio and proportion. If $\dfrac ab=\dfrac cd$ , then by componendo and dividendo

\frac{a+b}{a-b}=\frac{c+d}{c-d}

It also works backwards: from $\dfrac{a+b}{a-b}=k$ we get $\dfrac ab=\dfrac{k+1}{k-1}$ .

Means. For positive $a,b$ , $\dfrac{a+b}2\ge\sqrt{ab}$ , with equality only when $a=b$ . In particular $t+\dfrac1t\ge2$ for $t>0$ . The full theory is in the Relation Between AM, GM and HM note.

Example 6 (IITianForum). If $x+\dfrac1x=3$ , find $x^2+\dfrac1{x^2}$ , $x^3+\dfrac1{x^3}$ and $x^5+\dfrac1{x^5}$ .

▸ SHOW SOLUTION

Step 1. Square and cube

x^2+\frac1{x^2}=3^2-2=7

x^3+\frac1{x^3}=3^3-3\cdot3=18

Step 2. Multiply the two

\left(x^2+\frac1{x^2}\right)\left(x^3+\frac1{x^3}\right)

=x^5+\frac1{x^5}+x+\frac1x

x^5+\frac1{x^5}=7\cdot18-3=123

Answer: $7$ , $18$ , $123$

Example 7 (IITianForum). Solve

(x-2)^3+(x-3)^3+(5-2x)^3=0

▸ SHOW SOLUTION

Step 1. The three terms add to zero

(x-2)+(x-3)+(5-2x)=0

So the sum of cubes equals three times the product.

Step 2. Solve

3(x-2)(x-3)(5-2x)=0

Answer: $x=2,\ 3,\ \tfrac52$

Example 8 (IITianForum). Find

\sum_{n=1}^{10}\frac{n}{n^4+n^2+1}

▸ SHOW SOLUTION

Step 1. Factorise

n^4+n^2+1=(n^2+n+1)(n^2-n+1)

The difference of the two factors is $2n$ .

Step 2. Split

Let $u_n=n^2-n+1$ . Then $u_{n+1}=n^2+n+1$ , and

\frac{n}{n^4+n^2+1}=\frac12\left[\frac{1}{u_n}-\frac{1}{u_{n+1}}\right]

Step 3. Telescope

Adding from $n=1$ to $10$ , the middle terms cancel, leaving $u_1=1$ and $u_{11}=111$ :

\frac12\left[\frac11-\frac1{111}\right]=\frac{55}{111}

Answer: $\dfrac{55}{111}$

Example 9 (IITianForum). Solve

\frac{\sqrt{x+5}+\sqrt{x-5}}{\sqrt{x+5}-\sqrt{x-5}}=3

▸ SHOW SOLUTION

Step 1. Componendo and dividendo

\frac{\sqrt{x+5}}{\sqrt{x-5}}=\frac{3+1}{3-1}=2

Step 2. Square

x+5=4(x-5)\ \Rightarrow\ x=\tfrac{25}3

$x>5$ , so both roots are defined.

Answer: $x=\dfrac{25}3$

Example 10 (IITianForum). Find the number of real solutions of

9^x+25^x+1=3^x+5^x+15^x

▸ SHOW SOLUTION

Step 1. Substitute

Put $a=3^x$ , $b=5^x$ . Then $9^x=a^2$ , $25^x=b^2$ , $15^x=ab$ :

a^2+b^2+1-a-b-ab=0

Step 2. Sum of squares

The left side is $Q$ for the three numbers $a$ , $b$ , $1$ :

\tfrac12\left[(a-b)^2+(a-1)^2+(b-1)^2\right]=0

So $a=b=1$ , that is $3^x=5^x=1$ , giving $x=0$ .

Answer: one solution, $x=0$

Example 11 (IITianForum). Find the number of real solutions of

(x^{10}+1)(1+x^2+x^4+x^6+x^8)=10x^9

▸ SHOW SOLUTION

Step 1. Negative x and zero

For $x\le0$ the left side is positive and the right side is $\le0$ . No solution.

Step 2. Positive x

By AM $\ge$ GM, $x^{10}+1\ge2x^5$ . For the five terms $1,x^2,x^4,x^6,x^8$ :

1+x^2+x^4+x^6+x^8\ge5\sqrt[5]{x^{20}}=5x^4

Multiplying, the left side is $\ge10x^9$ .

Step 3. Equality

Equality needs equal terms in both: $x^{10}=1$ and $1=x^2=\cdots=x^8$ , so $x=1$ .

Answer: $1$

Polynomials

Dividing $P(x)$ by $D(x)$ gives a quotient $Q(x)$ and a remainder $R(x)$ with $\deg R<\deg D$ :

P(x)=D(x)\,Q(x)+R(x)

Result	Statement
remainder theorem	remainder of $P(x)$ by $x-a$ is $P(a)$
factor theorem	$x-a$ divides $P(x)$ exactly when $P(a)=0$
divisor a quadratic	remainder is $\alpha x+\beta$
$a^n-b^n$	divisible by $a-b$ for every $n$
$a^n+b^n$	divisible by $a+b$ for odd $n$
identity theorem	degree $\le n$ and zero at $n+1$ points means zero everywhere

The last two divisibility facts are the factor theorem: put $a=b$ in $a^n-b^n$ , and $a=-b$ in $a^n+b^n$ with $n$ odd.

Example 12 (IITianForum). $x-2$ is a factor of $P(x)=x^3+ax^2+bx+6$ , and $P(x)$ leaves remainder $3$ when divided by $x-3$ . Find $a$ and $b$ .

▸ SHOW SOLUTION

Step 1. Two conditions

P(2)=0:\ 4a+2b=-14

P(3)=3:\ 9a+3b=-30

Step 2. Solve

$2a+b=-7$ and $3a+b=-10$ give $a=-3$ , $b=-1$ .

Answer: $a=-3$ , $b=-1$

Example 13 (IITianForum). Find the remainder when $x^{100}$ is divided by $x^2-3x+2$ .

▸ SHOW SOLUTION

Step 1. Form of the remainder

$x^2-3x+2=(x-1)(x-2)$ , so

x^{100}=(x-1)(x-2)Q(x)+\alpha x+\beta

Step 2. Put x = 1 and x = 2

\alpha+\beta=1,\qquad 2\alpha+\beta=2^{100}

\alpha=2^{100}-1,\qquad \beta=2-2^{100}

Answer: $(2^{100}-1)x+2-2^{100}$

Example 14 (JEE Main 2026, 28 January Shift 1). Let

S=\{x^3+ax^2+bx+c\}

with $a,b,c\in\mathbb{N}$ and $a,b,c\le20$

be a set of polynomials. Then the number of polynomials in $S$ which are divisible by $x^2+2$ is

Tap an option to check your answer

(1)

20

(2)

6

(3)

120

(4)

10

▸ SHOW SOLUTION

Step 1. The quotient is linear

x^3+ax^2+bx+c=(x^2+2)(x+k)

=x^3+kx^2+2x+2k

Step 2. Compare coefficients

$a=k$ , $b=2$ , $c=2k$ . So $c$ is even and $a=\dfrac c2$ .

Step 3. Count

$c\in\{2,4,\ldots,20\}$ gives $a\in\{1,\ldots,10\}$ , all allowed. Ten polynomials.

Answer: option (4)

Example 15 (JEE Main 2026, 28 January Shift 2). Given below two statements:

Statement I: $25^{13}+20^{13}+8^{13}+3^{13}$ is divisible by $7$ .

Statement II: The integral part of $\left(7+4\sqrt3\right)^{25}$ is an odd number.

Tap an option to check your answer

(1)Both Statement I and Statement II are false.

(2)Both Statement I and Statement II are true.

(3)Statement I is false but Statement II is true.

(4)Statement I is true but Statement II is false.

▸ SHOW SOLUTION

Step 1. Statement I

$13$ is odd, so $a+b$ divides $a^{13}+b^{13}$ . Pair the terms:

28\mid 25^{13}+3^{13},\qquad 28\mid 20^{13}+8^{13}

So $28$ , and hence $7$ , divides the sum. True.

Step 2. Statement II: the conjugate

Let $R=(7+4\sqrt3)^{25}=I+f$ with $0\le f<1$ , and $R'=(7-4\sqrt3)^{25}$ . Since $0<7-4\sqrt3<1$ , $0<R'<1$ .

Step 3. R + R′ is an even integer

In the expansions of $(7+4\sqrt3)^{25}$ and $(7-4\sqrt3)^{25}$ , the terms with odd powers of $4\sqrt3$ cancel and the rest double. Each even power of $4\sqrt3$ is an integer, so $R+R'$ is even.

Step 4. Conclude

$I+f+R'$ is an integer and $0<f+R'<2$ , so $f+R'=1$ . Then $I=(R+R')-1$ is odd. True.

Answer: option (2)

Example 16 (JEE Main 2025, 24 January Shift 1). The product of all the rational roots of the equation

(x^2-9x+11)^2-(x-4)(x-5)=3

is equal to

Tap an option to check your answer

(1)

14

(2)

7

(3)

28

(4)

21

▸ SHOW SOLUTION

Step 1. One substitution

$(x-4)(x-5)=x^2-9x+20$ . Put $t=x^2-9x$ :

(t+11)^2-(t+20)=3

t^2+21t+98=0\ \Rightarrow\ t=-7,\ -14

Step 2. Back to x

$t=-14$ : $x^2-9x+14=0$ , so $x=2,7$ .

$t=-7$ : $x^2-9x+7=0$ , discriminant $53$ , not a square, so irrational roots.

Step 3. Product

$2\times7=14$ .

Answer: option (1)

Example 17 (IITianForum). $P(x)$ is a polynomial of degree $5$ with

P(k)=\frac{k}{k+1},\qquad k=0,1,2,3,4,5

Find $P(6)$ .

▸ SHOW SOLUTION

Step 1. Build a polynomial with known roots

Let $Q(x)=(x+1)P(x)-x$ . It has degree $6$ and is zero at $x=0,1,\ldots,5$ :

Q(x)=c\,x(x-1)(x-2)\cdots(x-5)

Step 2. Find c

Put $x=-1$ . The left side is $0\cdot P(-1)+1=1$ :

1=c\,(-1)(-2)\cdots(-6)=720c

So $c=\dfrac1{720}$ .

Step 3. Put x = 6

7P(6)-6=\frac{6!}{720}=1

Answer: $P(6)=1$

Inequalities

Step	Rule
add or subtract $k$	direction stays
multiply by $k>0$	direction stays
multiply by $k<0$	direction reverses
squaring	allowed only when both sides are $\ge0$
reciprocals, same sign	direction reverses

Never cross-multiply by an unknown sign

In $\dfrac{P(x)}{Q(x)}>k$ , multiplying by $Q(x)$ is wrong unless the sign of $Q(x)$ is known. Bring everything to one side instead.

The wavy curve method

Bring everything to one side: $f(x)>0$ , $<0$ , $\ge0$ or $\le0$ .
Factorise completely. Write every linear factor as $x-a$ ; take the minus signs out to the front.
Mark the roots in increasing order on a number line.
With a positive leading coefficient, start above the line on the far right and alternate through each root.
Read off the intervals with the required sign.

A number line with the points minus 2, 1, 3 and 5 marked; a wave starts above the line on the far right and crosses the line at each point, with plus and minus signs written alternately in the five gaps, plus on the far right

Example 18 (IITianForum). Solve

(x+2)(x-1)(3-x)(x-5)\le0

▸ SHOW SOLUTION

Step 1. Make every factor x − a

$3-x=-(x-3)$ . Multiply by $-1$ and reverse the sign:

(x+2)(x-1)(x-3)(x-5)\ge0

Step 2. Wave

Roots $-2,1,3,5$ ; signs from the right: $+,-,+,-,+$ . Take the $+$ parts with the roots included.

Answer: $(-\infty,-2]\cup[1,3]\cup[5,\infty)$

Repeated factors and rational inequalities

An odd power of $x-a$ changes the sign at $a$ : the wave crosses.
An even power does not: the wave touches the line and turns back. The root still goes on the line.
A quadratic factor with positive leading coefficient and negative discriminant, like $x^2+x+1$ , is always positive and can be dropped.
For $\dfrac{P}{Q}$ , use the same signs as $PQ$ . Roots of $Q$ are always hollow; roots of $P$ are filled only for $\ge$ or $\le$ .

Example 19 (IITianForum). Solve

\frac{(x-1)^2(x+2)^3}{(x-4)(x+3)^2}\ge0

▸ SHOW SOLUTION

Step 1. Mark the roots

Numerator: $1$ (even, filled), $-2$ (odd, filled). Denominator: $-3$ (even, hollow), $4$ (odd, hollow).

Step 2. Wave

From the right: $+$ on $(4,\infty)$ ; crossing at $4$ , $-$ on $(1,4)$ ; touching at $1$ , still $-$ on $(-2,1)$ ; crossing at $-2$ , $+$ on $(-3,-2)$ ; touching at $-3$ , still $+$ on $(-\infty,-3)$ .

Step 3. Collect

The $+$ intervals, the filled root $-2$ , and the point $1$ where the expression is zero.

Answer: $(-\infty,-3)\cup(-3,-2]\cup\{1\}\cup(4,\infty)$

A number line with the points minus 3, minus 2, 1 and 4; hollow dots at minus 3 and 4, filled dots at minus 2 and 1; a wave that starts above the line on the far right, crosses at 4, touches the line at 1 and turns back, crosses at minus 2, and touches at minus 3 staying above

Example 20 (IITianForum). Solve $\dfrac{2x-3}{x+1}\ge1$ .

▸ SHOW SOLUTION

Step 1. One side

\frac{2x-3}{x+1}-1\ge0\ \Rightarrow\ \frac{x-4}{x+1}\ge0

Step 2. Wave

Roots $-1$ (hollow) and $4$ (filled); $+$ outside them.

Answer: $(-\infty,-1)\cup[4,\infty)$

Cross-multiplying would give $2x-3\ge x+1$ , that is $x\ge4$ only, and loses $(-\infty,-1)$ , where $x+1<0$ .

Inequalities with a square root

A square root is defined only for a non-negative argument, and it is never negative.

Inequality	Equivalent system
$\sqrt f<g$	$f\ge0$ , $g>0$ , $f<g^2$
$\sqrt f>g$	$f\ge0$ with $g<0$ ; or $g\ge0$ with $f>g^2$
$\sqrt f<\sqrt g$	$f\ge0$ , $f<g$

Example 21 (IITianForum). Solve $\sqrt{x+3}>x+1$ .

▸ SHOW SOLUTION

Step 1. Case x + 1 < 0

The left side is $\ge0$ , so the inequality holds wherever the root is defined: $-3\le x<-1$ .

Step 2. Case x + 1 ≥ 0

Square:

x+3>(x+1)^2\ \Rightarrow\ x^2+x-2<0

$-2<x<1$ ; with $x\ge-1$ this gives $-1\le x<1$ .

Step 3. Union

Answer: $[-3,1)$

Example 22 (IITianForum). Solve $\sqrt{x^2-3x-10}<8-x$ .

▸ SHOW SOLUTION

Step 1. Conditions

$x^2-3x-10=(x-5)(x+2)\ge0$ : $x\le-2$ or $x\ge5$ . Also $8-x>0$ : $x<8$ .

Step 2. Square

x^2-3x-10<64-16x+x^2

13x<74\ \Rightarrow\ x<\tfrac{74}{13}

Step 3. Intersect

Answer: $(-\infty,-2]\cup\left[5,\tfrac{74}{13}\right)$

Modulus

|x|=x\ \ (x\ge0),\qquad |x|=-x\ \ (x<0)

$|x|$ is the distance of $x$ from $0$ , and $|x-a|$ is the distance between $x$ and $a$ on the number line.

Property	Statement
sign	$\lvert x\rvert\ge0$ ; zero only at $x=0$
square	$\lvert x\rvert^2=x^2$ , $\sqrt{x^2}=\lvert x\rvert$
product	$\lvert xy\rvert=\lvert x\rvert\lvert y\rvert$
quotient	$\lvert x/y\rvert=\lvert x\rvert/\lvert y\rvert$ , $y\ne0$

The square root of a square

$\sqrt{x^2}=|x|$ , not $x$ . For example, for $-2\le x\le3$ , $\sqrt{(x-3)^2}+\sqrt{(x+2)^2}=(3-x)+(x+2)=5$ .

Modulus equations

Equation	Method
$\lvert f\rvert=a$ , $a>0$	$f=a$ or $f=-a$
$\lvert f\rvert=g$	$f=\pm g$ , keep roots with $g\ge0$
$\lvert f\rvert=\lvert g\rvert$	$f=g$ or $f=-g$
$(x-a)^2$ and $\lvert x-a\rvert$ together	put $t=\lvert x-a\rvert\ge0$
several moduli	split at the zeros, one case per interval

Example 23 (JEE Main 2026, 21 January Shift 1). The sum of all the roots of the equation $(x-1)^2-5|x-1|+6=0$ is

Tap an option to check your answer

(1)

4

(2)

3

(3)

1

(4)

5

▸ SHOW SOLUTION

Step 1. Substitute

$(x-1)^2=|x-1|^2$ . Put $t=|x-1|\ge0$ :

t^2-5t+6=0\ \Rightarrow\ t=2,\ 3

Step 2. Roots

$|x-1|=2$ : $x=3,-1$ . $|x-1|=3$ : $x=4,-2$ .

3-1+4-2=4

Answer: option (1)

Example 24 (JEE Main 2026, 24 January Shift 1). The number of real solutions of the equation $x|x+3|+|x-1|-2=0$ is

Tap an option to check your answer

(1)

3

(2)

2

(3)

5

(4)

4

▸ SHOW SOLUTION

Step 1. Case x ≥ 1

x^2+4x-3=0\ \Rightarrow\ x=-2\pm\sqrt7

$-2+\sqrt7\approx0.65$ and $-2-\sqrt7$ are both below $1$ . No root.

Step 2. Case −3 ≤ x < 1

x^2+2x-1=0\ \Rightarrow\ x=-1\pm\sqrt2

$0.41$ and $-2.41$ both lie in $[-3,1)$ . Two roots.

Step 3. Case x < −3

x^2+4x+1=0\ \Rightarrow\ x=-2\pm\sqrt3

Only $-2-\sqrt3\approx-3.73$ is below $-3$ . One root.

Answer: $3$ roots, option (1)

The graph of y equals x times the modulus of x plus 3, plus the modulus of x minus 1, drawn as a smooth curve made of three pieces, cut by the horizontal line y equals 2 at three marked points near x equals minus 3.7, minus 2.4 and 0.4

Example 25 (JEE Main 2025, 24 January Shift 2). Let

A=\left\{x\in(0,\pi)-\left\{\tfrac\pi2\right\}:\ L=2\right\}

L=\log_{(2/\pi)}|\sin x|+\log_{(2/\pi)}|\cos x|

B=\left\{x\ge0:\ M=0\right\}

M=\sqrt x(\sqrt x-4)-3|\sqrt x-2|+6

Then $n(A\cup B)$ is equal to

Tap an option to check your answer

(1)

4

(2)

2

(3)

8

(4)

6

▸ SHOW SOLUTION

Step 1. Set B: a hidden square

Put $t=\sqrt x\ge0$ . Then $t^2-4t+6=(t-2)^2+2$ :

(t-2)^2-3|t-2|+2=0

|t-2|=1\ \text{or}\ 2\ \Rightarrow\ t=1,3,0,4

$B=\{0,1,9,16\}$ .

Step 2. Set A

|\sin x\cos x|=\left(\tfrac2\pi\right)^2\ \Rightarrow\ |\sin2x|=\tfrac8{\pi^2}

$\tfrac8{\pi^2}\approx0.81$ . As $x$ runs over $(0,\pi)$ , $2x$ runs over $(0,2\pi)$ , where $\sin2x=\pm0.81$ has $4$ solutions.

Step 3. Union

Of the points of $B$ , only $1$ lies in $(0,\pi)$ , and $|\sin2|\approx0.91\ne0.81$ . So $1\notin A$ and the sets do not overlap.

n(A\cup B)=4+4=8

Answer: option (3)

Modulus inequalities

For $a>0$ :

|x|<a\iff -a<x<a

|x|>a\iff x<-a\ \text{or}\ x>a

Two number lines: on the upper one the segment between minus a and a is highlighted with hollow end points; on the lower one the two rays to the left of minus a and to the right of a are highlighted, with hollow end points

In the same way, $|f|<g$ means $-g<f<g$ , and $|f|>g$ means $f>g$ or $f<-g$ .

Example 26 (JEE Main 2026, 21 January Shift 2). Let $A=\{x:|x^2-10|\le6\}$ and $B=\{x:|x-2|>1\}$ . Then

Tap an option to check your answer

(1)

A\cup B=(-\infty,1]\cup(2,\infty)

(2)

A-B=[2,3)

(3)

A\cap B=[-4,-2]\cup[3,4]

(4)

B-A=(-\infty,-4)\cup(-2,1)\cup(4,\infty)

▸ SHOW SOLUTION

Step 1. Set A

-6\le x^2-10\le6\ \Rightarrow\ 4\le x^2\le16

$A=[-4,-2]\cup[2,4]$ .

Step 2. Set B

$x-2>1$ or $x-2<-1$ : $B=(-\infty,1)\cup(3,\infty)$ .

Step 3. B − A

From $(-\infty,1)$ remove $[-4,-2]$ ; from $(3,\infty)$ remove $(3,4]$ :

B-A=(-\infty,-4)\cup(-2,1)\cup(4,\infty)

The others fail: $A\cap B=[-4,-2]\cup(3,4]$ , $A-B=[2,3]$ , $A\cup B=(-\infty,1)\cup[2,\infty)$ .

Answer: option (4)

Several moduli

Find the critical points, where each expression inside a modulus is zero.
On each interval between them, open every modulus with a fixed sign.
Solve, and keep only the part of the answer inside that interval.
Take the union.

Example 27 (IITianForum). Solve $|x-1|+|x-3|<6$ .

▸ SHOW SOLUTION

Step 1. Case x < 1

(1-x)+(3-x)<6\ \Rightarrow\ x>-1

So $-1<x<1$ .

Step 2. Case 1 ≤ x < 3

$(x-1)+(3-x)=2<6$ , true for the whole interval.

Step 3. Case x ≥ 3

(x-1)+(x-3)<6\ \Rightarrow\ x<5

So $3\le x<5$ .

Answer: $(-1,5)$

As distances: the sum of the distances from $1$ and from $3$ is $2$ between them and grows by $2$ for each unit outside, reaching $6$ at $-1$ and $5$ .

The graph of y equals the modulus of x minus 1 plus the modulus of x minus 3: a flat bottom at height 2 between x equals 1 and x equals 3, rising lines on both sides, cut by the dashed line y equals 6 at x equals minus 1 and x equals 5, with the part of the x axis between them highlighted

The triangle inequality

\big||x|-|y|\big|\le|x\pm y|\le|x|+|y|

$|x+y|=|x|+|y|$ exactly when $xy\ge0$ , and $|x-y|=|x|+|y|$ exactly when $xy\le0$ .

Example 28 (IITianForum). Solve $|2x-1|+|x+3|=|3x+2|$ .

▸ SHOW SOLUTION

Step 1. Spot the sum

$(2x-1)+(x+3)=3x+2$ , so the equation is $|a|+|b|=|a+b|$ with $a=2x-1$ , $b=x+3$ .

Step 2. Equality case

(2x-1)(x+3)\ge0

Answer: $(-\infty,-3]\cup\left[\tfrac12,\infty\right)$

Nested modulus and counting by graph

For $|f(x)|$ , reflect the parts of $y=f(x)$ below the $x$ axis. The number of solutions of $f(x)=k$ is the number of meets of $y=f(x)$ with the line $y=k$ .

Example 29 (IITianForum). The equation $\big||x-1|-a\big|=2$ has exactly three distinct real solutions. Find $a$ and the solutions.

▸ SHOW SOLUTION

Step 1. Open the outer modulus

|x-1|=a+2\quad\text{or}\quad |x-1|=a-2

Step 2. Count

$|x-1|=c$ has two solutions for $c>0$ , one for $c=0$ , none for $c<0$ . Three solutions in all needs $a+2>0$ and $a-2=0$ .

Step 3. Solve with a = 2

$|x-1|=4$ : $x=5,-3$ . $|x-1|=0$ : $x=1$ .

Answer: $a=2$ ; $x=-3,1,5$

The graph of y equals the modulus of the modulus of x minus 1, minus 2: a W shape with zeros at x equals minus 1 and x equals 3 and a peak of height 2 at x equals 1, cut by the dashed line y equals 2 at three marked points x equals minus 3, 1 and 5

Example 30 (JEE Main 2026, 23 January Shift 2). Consider two sets $A=\{x\in\mathbb{Z}:\ \big||x-3|-3\big|\le1\}$ and

B=\left\{x\in\mathbb{R}-\{1,2\}:\ g(x)=0\right\}

g(x)=\frac{(x-2)(x-4)}{x-1}\log_e|x-2|

Then the number of onto functions $f:A\to B$ is equal to

Tap an option to check your answer

(1)

62

(2)

79

(3)

32

(4)

81

▸ SHOW SOLUTION

Step 1. Set A

-1\le|x-3|-3\le1

2\le|x-3|\le4

$x-3\in[2,4]$ or $x-3\in[-4,-2]$ : $x\in[5,7]\cup[-1,1]$ . Integers: $A=\{-1,0,1,5,6,7\}$ .

Step 2. Set B

$x-4=0$ gives $x=4$ . $\log_e|x-2|=0$ gives $|x-2|=1$ , so $x=3$ (and $x=1$ , excluded). $B=\{3,4\}$ .

Step 3. Onto functions

All functions minus the two constant ones:

2^6-2=62

Answer: option (1)

Example 31 (JEE Main 2025, 24 January Shift 2). The number of real solution(s) of the equation

x^2+3x+2=\min\{|x-3|,\ |x+2|\}

Tap an option to check your answer

(1)

2

(2)

0

(3)

3

(4)

1

▸ SHOW SOLUTION

Step 1. The right side

$|x-3|$ and $|x+2|$ are equal at $x=\tfrac12$ . So

\min=|x+2|\ \ \text{for } x\le\tfrac12

\min=|x-3|\ \ \text{for } x\ge\tfrac12

Step 2. x ≤ 1/2

$(x+1)(x+2)=|x+2|$ . Clearly $x=-2$ is a root. For $-2<x\le\tfrac12$ , divide by $x+2>0$ : $x+1=1$ , so $x=0$ . For $x<-2$ , divide by $-(x+2)>0$ : $x+1=-1$ , so $x=-2$ , which is not below $-2$ .

Step 3. x ≥ 1/2

For $\tfrac12\le x\le3$ : $x^2+3x+2=3-x$ gives $x^2+4x-1=0$ , $x=-2\pm\sqrt5$ ; the larger root $0.236$ is below $\tfrac12$ . For $x>3$ : $x^2+2x+5=0$ has no real root.

Answer: $x=-2,0$ ; option (1)

An upward parabola y equals x squared plus 3x plus 2 and the graph of the minimum of the moduli of x minus 3 and x plus 2, made of straight pieces that touch the x axis at x equals minus 2 and x equals 3 with a peak at x equals one half; the two graphs meet at two marked points, x equals minus 2 and x equals 0

Greatest integer and fractional part

$[x]$ is the greatest integer $\le x$ , and $\{x\}=x-[x]\in[0,1)$ . So $[2.7]=2$ and $[-2.7]=-3$ .

[x]\le x<[x]+1

An equation in $[x]$ is solved for the integer $n=[x]$ first; then $x\in[n,n+1)$ . The graphs of $[x]$ and $\{x\}$ are in the Functions note.

Example 32 (IITianForum). Solve (a) $[x]^2-5[x]+6=0$ ; (b) $4\{x\}=x+[x]$ .

▸ SHOW SOLUTION

Step 1. Part (a)

$([x]-2)([x]-3)=0$ , so $[x]=2$ or $3$ . Then $x\in[2,3)\cup[3,4)=[2,4)$ .

Step 2. Part (b)

Write $x=n+f$ with $n=[x]$ , $f=\{x\}$ :

4f=2n+f\ \Rightarrow\ f=\tfrac{2n}3

$0\le f<1$ allows $n=0,1$ , giving $f=0,\tfrac23$ .

Answer: (a) $[2,4)$ ; (b) $x=0,\ \tfrac53$

Logarithms

\log_a N=x\iff a^x=N

The logarithm exists only when

N>0,\qquad a>0,\qquad a\ne1

$\log N$ means $\log_{10}N$ and $\ln N$ means $\log_e N$ . Always $\log_a1=0$ , $\log_a a=1$ and $a^{\log_a N}=N$ .

Example 33 (IITianForum). (a) Find $\log_2\log_3 81$ , $3^{\log_9 25}$ and $\log_{\sqrt2}8$ . (b) For which $x$ is $\log_{x-1}(5-x)$ defined?

▸ SHOW SOLUTION

Step 1. Part (a)

\log_3 81=4,\quad \log_2 4=2

3^{\log_9 25}=3^{\log_3 5}=5

\log_{\sqrt2}8=\log_{2^{1/2}}2^3=6

Step 2. Part (b)

Argument $5-x>0$ ; base $x-1>0$ and $x-1\ne1$ :

x<5,\qquad x>1,\qquad x\ne2

Answer: (a) $2$ , $5$ , $6$ ; (b) $(1,2)\cup(2,5)$

Laws of logarithms

For valid bases and positive $M$ , $N$ :

Law	Statement
product	$\log_a(MN)=\log_a M+\log_a N$
quotient	$\log_a\frac MN=\log_a M-\log_a N$
power	$\log_a M^p=p\log_a M$
base power	$\log_{a^q}M^p=\frac pq\log_a M$

The power law can change the domain

$\log_a x^2$ is defined for all $x\ne0$ , but $2\log_a x$ only for $x>0$ . The correct form is $\log_a x^2=2\log_a|x|$ .

Example 34 (IITianForum). Solve (a) $\log_3(x-1)^2=2$ ; (b) $2\log_3(x-1)=2$ .

▸ SHOW SOLUTION

Step 1. Part (a)

$(x-1)^2=9$ , so $x-1=\pm3$ : $x=4$ or $x=-2$ . Both make $(x-1)^2>0$ .

Step 2. Part (b)

Here $x-1>0$ is needed: $\log_3(x-1)=1$ , $x=4$ .

Answer: (a) $x=4,-2$ ; (b) $x=4$

Example 35 (IITianForum). Let $\log_{1/9}\beta=-1$ , $\log_\beta\alpha=-1$ and $\log_\alpha 27=\gamma$ . Find

\left(\frac1\alpha+1\right)^{\log_{\sqrt{10}}\left(\frac{\beta^2}{81}+4\gamma^2\right)}

▸ SHOW SOLUTION

Step 1. The three values

\beta=\left(\tfrac19\right)^{-1}=9,\qquad \alpha=9^{-1}=\tfrac19

\gamma=\log_{3^{-2}}3^3=-\tfrac32

Step 2. The exponent

\frac{\beta^2}{81}+4\gamma^2=1+9=10

\log_{\sqrt{10}}10=2

Step 3. Value

\left(9+1\right)^2=100

Answer: $100$

Example 36 (IITianForum). If $a,b>0$ and

\ln\frac{a+b}{4}=\frac{\ln a+\ln b}{2},

find $\dfrac ab+\dfrac ba$ .

▸ SHOW SOLUTION

Step 1. Remove the logs

The right side is $\ln\sqrt{ab}$ , so

\frac{a+b}4=\sqrt{ab}\ \Rightarrow\ (a+b)^2=16ab

Step 2. Divide by ab

a^2+b^2=14ab\ \Rightarrow\ \frac ab+\frac ba=14

Answer: $14$

Change of base

\log_a N=\frac{\log_b N}{\log_b a}

Consequence	Statement
reciprocal	$\log_a b\cdot\log_b a=1$
chain	$\log_a b\cdot\log_b c\cdot\log_c d=\log_a d$
swap	$a^{\log_b c}=c^{\log_b a}$

Example 37 (IITianForum). Let $2^{a_1}=3$ , $3^{a_2}=4$ , $4^{a_3}=5$ , and so on up to $15^{a_{14}}=16$ . Find $a_1a_2\cdots a_{14}$ .

▸ SHOW SOLUTION

Step 1. Each term is a log

$a_1=\log_2 3$ , $a_2=\log_3 4$ , $\ldots$ , $a_{14}=\log_{15}16$ .

Step 2. The chain

a_1a_2\cdots a_{14}=\log_2 16=4

Answer: $4$

Example 38 (IITianForum). Find the smallest integer greater than

\frac{1}{\log_2\pi}+\frac{1}{\log_5\pi}

▸ SHOW SOLUTION

Step 1. Flip the bases

\frac1{\log_2\pi}+\frac1{\log_5\pi}=\log_\pi 2+\log_\pi 5

=\log_\pi 10

Step 2. Place 10 between powers of π

$\pi^2\approx9.87<10<\pi^3\approx31$ , so $2<\log_\pi10<3$ .

Answer: $3$

Example 39 (JEE Main 2026, 4 April Shift 2). Let $\alpha=\dfrac14+\dfrac18+\dfrac1{16}+\cdots\infty$ and $\beta=\dfrac13+\dfrac19+\dfrac1{27}+\cdots\infty$ . Then the value of

(0.2)^{\log_{\sqrt5}\alpha}+(0.04)^{\log_5\beta}

Tap an option to check your answer

(1)

4

(2)

5

(3)

8

(4)

25

▸ SHOW SOLUTION

Step 1. The two sums

\alpha=\frac{1/4}{1-1/2}=\frac12,\qquad \beta=\frac{1/3}{1-1/3}=\frac12

Step 2. First term by the swap rule

$\log_{\sqrt5}0.2=\log_{5^{1/2}}5^{-1}=-2$ :

(0.2)^{\log_{\sqrt5}\alpha}=\alpha^{\log_{\sqrt5}0.2}=\alpha^{-2}=4

Step 3. Second term

(0.04)^{\log_5\beta}=5^{-2\log_5\beta}=\beta^{-2}=4

Answer: $4+4=8$ , option (3)

Graph of the logarithm

Two graphs side by side: on the left the increasing curve y equals log to base a of x for a greater than 1, and on the right the decreasing curve y equals log to base a of x for a between 0 and 1; both pass through the marked point 1 comma 0 and approach the y axis without touching it

Base	Behaviour
$a>1$	increasing; $\log_a x<0$ on $(0,1)$ , $>0$ on $(1,\infty)$
$0<a<1$	decreasing; $\log_a x>0$ on $(0,1)$ , $<0$ on $(1,\infty)$

So $\log_a x>0$ exactly when $x$ and $a$ lie on the same side of $1$ . This gives the sign of a log at a glance.

Logarithmic equations

Write the domain: every argument positive, every base positive and not $1$ .
Bring the logs to one base and combine them.
Remove the log: $\log_a M=\log_a N$ gives $M=N$ .
Reject roots outside the domain.

Why roots must be checked

$\log M+\log N$ needs $M>0$ and $N>0$ , but $\log(MN)$ needs only $MN>0$ . Combining the logs enlarges the domain, and the extra root that appears is the one the domain rejects.

Example 40 (IITianForum). Solve $\log_3(x-2)+\log_3(x+4)=3$ .

▸ SHOW SOLUTION

Step 1. Domain

$x-2>0$ and $x+4>0$ : $x>2$ .

Step 2. Combine

(x-2)(x+4)=27

x^2+2x-35=0

$x=5$ or $x=-7$ .

Step 3. Check

$-7$ is outside the domain.

Answer: $x=5$

Example 41 (JEE Main 2025, 22 January Shift 1). The product of all solutions of the equation $e^{5(\log_e x)^2+3}=x^8$ , $x>0$ , is

Tap an option to check your answer

(1)

e^{8/5}

(2)

e^{6/5}

(3)

e^2

(4)

e

▸ SHOW SOLUTION

Step 1. Take ln

Put $t=\ln x$ :

5t^2+3=8t\ \Rightarrow\ 5t^2-8t+3=0

Step 2. Product of the x values

$x_1x_2=e^{t_1+t_2}$ and $t_1+t_2=\tfrac85$ . (The roots $t=1,\tfrac35$ are real and distinct.)

Answer: $e^{8/5}$ , option (1)

Example 42 (JEE Main 2026, 8 April Shift 2). The sum of squares of all real solutions of

\log_{(x+1)}(2x^2+5x+3)

=4-\log_{(2x+3)}(x^2+2x+1)

is ______.

▸ SHOW SOLUTION

Step 1. Factorise the arguments

$2x^2+5x+3=(2x+3)(x+1)$ and $x^2+2x+1=(x+1)^2$ . Bases need $x+1>0$ , $x+1\ne1$ , $2x+3>0$ , $2x+3\ne1$ ; so $x>-1$ , $x\ne0$ .

Step 2. One variable

Let $t=\log_{x+1}(2x+3)$ . Then $\log_{2x+3}(x+1)=\tfrac1t$ :

1+t+\frac2t=4\ \Rightarrow\ t^2-3t+2=0

$t=1$ or $t=2$ .

Step 3. Back to x

$t=1$ : $2x+3=x+1$ , $x=-2$ , rejected.

$t=2$ : $2x+3=(x+1)^2$ , $x^2=2$ ; only $x=\sqrt2$ is in the domain.

Answer: $(\sqrt2)^2=2$

Example 43 (JEE Main 2026, 23 January Shift 2). The sum of all the real solutions of the equation

\log_{(x+3)}(6x^2+28x+30)

=5-2\log_{(6x+10)}(x^2+6x+9)

is equal to

Tap an option to check your answer

(1)

2

(2)

1

(3)

0

(4)

4

▸ SHOW SOLUTION

Step 1. Factorise

$6x^2+28x+30=(x+3)(6x+10)$ and $x^2+6x+9=(x+3)^2$ .

Step 2. One variable

Let $A=\log_{x+3}(6x+10)$ :

1+A=5-\frac4A\ \Rightarrow\ (A-2)^2=0

Step 3. Back to x

$6x+10=(x+3)^2$ gives $x^2=1$ , $x=\pm1$ . Bases $4,2$ and $16,4$ : all valid.

Answer: $1+(-1)=0$ , option (3)

Example 44 (IITianForum). Find the number of solutions of

\log_3(x^2+2)=\log_{1/2}\left(x+\frac1x\right)

for $x>0$ .

▸ SHOW SOLUTION

Step 1. Left side

$x^2+2\ge2$ , so the left side is $\ge\log_3 2>0$ .

Step 2. Right side

$x+\dfrac1x\ge2$ and the base $\tfrac12$ is below $1$ , so the right side is $\le\log_{1/2}2=-1$ .

Step 3. Compare

A positive number never equals a number $\le-1$ .

Answer: no solution

For x greater than 0, the curve y equals log to base 3 of x squared plus 2, which stays above the dashed level log to base 3 of 2, and the curve y equals log to base one half of x plus 1 over x, which stays below the dashed level minus 1, with its highest point at x equals 1; the two curves never meet

Logarithmic inequalities

For $a>1$ :

\log_a M>\log_a N\iff M>N>0

For $0<a<1$ :

\log_a M>\log_a N\iff 0<M<N

The decreasing curve y equals log to base one half of u for u greater than 0, cut by the dashed line y equals minus 2 at u equals 4; the part of the curve above the line lies over the interval from 0 to 4, which is highlighted on the u axis

If the base contains the variable, solve the two cases $a>1$ and $0<a<1$ separately and take the union.

Example 45 (IITianForum). Solve $\log_{1/2}(x-1)>-2$ .

▸ SHOW SOLUTION

Step 1. Write −2 as a log

$-2=\log_{1/2}4$ . The base is below $1$ , so the direction reverses:

0<x-1<4

Answer: $(1,5)$

Example 46 (IITianForum). Solve $\log_x(x+2)>2$ .

▸ SHOW SOLUTION

Step 1. Case x > 1

x+2>x^2\ \Rightarrow\ (x-2)(x+1)<0

$-1<x<2$ ; with $x>1$ : $1<x<2$ .

Step 2. Case 0 < x < 1

0<x+2<x^2\ \Rightarrow\ (x-2)(x+1)>0

$x>2$ or $x<-1$ : nothing in $(0,1)$ .

Answer: $(1,2)$

Example 47 (JEE Main 2025, 29 January Shift 2). If the domain of the function $\log_5(18x-x^2-77)$ is $(\alpha,\beta)$ and the domain of the function

\log_{(x-1)}\left(\frac{2x^2+3x-2}{x^2-3x-4}\right)

is $(\gamma,\delta)$ , then $\alpha^2+\beta^2+\gamma^2$ is equal to

Tap an option to check your answer

(1)

195

(2)

174

(3)

186

(4)

179

▸ SHOW SOLUTION

Step 1. First domain

x^2-18x+77<0

(x-7)(x-11)<0

$(\alpha,\beta)=(7,11)$ .

Step 2. Second domain: the argument

\frac{(2x-1)(x+2)}{(x-4)(x+1)}>0

Wave with roots $-2,-1,\tfrac12,4$ : $x<-2$ , $-1<x<\tfrac12$ or $x>4$ .

Step 3. The base

$x>1$ , $x\ne2$ . With Step 2: $x>4$ , so $\gamma=4$ .

Answer: $49+121+16=186$ , option (3)

Example 48 (JEE Main 2026, 2 April Shift 1). If the domain of the function

f(x)=\sqrt{\log_{(0.6)}\left|\frac{2x-5}{x^2-4}\right|}

is $(-\infty,a]\cup\{b\}\cup[c,d)\cup(e,\infty)$ , then the value of $a+b+c+d+e$ is ______.

▸ SHOW SOLUTION

Step 1. Remove the log

The root needs the log $\ge0$ . The base $0.6$ is below $1$ , so

0<\left|\frac{2x-5}{x^2-4}\right|\le1

That is $x\ne\pm2$ , $x\ne\tfrac52$ , and $-1\le\dfrac{2x-5}{x^2-4}\le1$ .

Step 2. Lower half

\frac{x^2+2x-9}{x^2-4}\ge0

Roots $-1\pm\sqrt{10}$ and $\pm2$ : $x\le-1-\sqrt{10}$ , $-2<x<2$ or $x\ge-1+\sqrt{10}$ .

Step 3. Upper half

\frac{(x-1)^2}{(x-2)(x+2)}\ge0

The square touches at $1$ : $x<-2$ , $x=1$ or $x>2$ .

Step 4. Intersect

(-\infty,-1-\sqrt{10}\,]\cup\{1\}

\cup\,[-1+\sqrt{10},\tfrac52)\cup(\tfrac52,\infty)

$a+c=-2$ , $b=1$ , $d=e=\tfrac52$ .

Answer: $-2+1+5=4$

Example 49 (IITianForum). Solve

\sqrt{\log_3x-1}-\log_3x+3>0

▸ SHOW SOLUTION

Step 1. Substitute

Put $t=\log_3 x$ . The root needs $t\ge1$ :

\sqrt{t-1}>t-3

Step 2. Case t < 3

The right side is negative: true for $1\le t<3$ .

Step 3. Case t ≥ 3

t-1>(t-3)^2\ \Rightarrow\ (t-2)(t-5)<0

With $t\ge3$ : $3\le t<5$ .

Step 4. Back to x

$1\le\log_3 x<5$ .

Answer: $[3,243)$

Characteristic, mantissa and digits

Write $\log_{10}N=$ characteristic $+$ mantissa, with the characteristic an integer and the mantissa in $[0,1)$ .

Result	Statement
digits in $N\ge1$	characteristic $+1$
zeros after the point, $0<N<1$	$\lvert$ characteristic $\rvert-1$
useful values	$\log2\approx0.3010$ , $\log3\approx0.4771$ , $\log7\approx0.8451$

Example 50 (IITianForum). (a) How many digits does $2^{100}$ have, and what is its first digit? (b) How many zeros come right after the decimal point in $3^{-20}$ ?

▸ SHOW SOLUTION

Step 1. Part (a)

\log 2^{100}=30.10

Characteristic $30$ : there are $31$ digits. The mantissa $0.10$ lies between $\log1=0$ and $\log2=0.301$ , so the first digit is $1$ .

Step 2. Part (b)

\log3^{-20}=-9.542=-10+0.458

Characteristic $-10$ : there are $9$ zeros after the point.

Answer: (a) $31$ digits, first digit $1$ ; (b) $9$

Exponential equations

$a^{f(x)}=a^{g(x)}$ with $a>0$ , $a\ne1$ gives $f(x)=g(x)$ .
$a^{2x}+ka^x+m=0$ is a quadratic in $t=a^x$ , and $t>0$ is compulsory.
$a^x>a^y$ means $x>y$ for $a>1$ , and $x<y$ for $0<a<1$ .

Example 51 (IITianForum). Solve $4^x+2^{x+2}=32$ .

▸ SHOW SOLUTION

Step 1. Substitute

Put $t=2^x>0$ :

t^2+4t-32=0\ \Rightarrow\ t=4\ \text{or}\ t=-8

Step 2. Reject

$t=-8$ is not allowed. $2^x=4$ .

Answer: $x=2$

Example 52 (IITianForum). Solve $3^x+4^x=5^x$ .

▸ SHOW SOLUTION

Step 1. Divide by 5^x

\left(\tfrac35\right)^x+\left(\tfrac45\right)^x=1

Step 2. One root

Both bases are below $1$ , so the left side decreases as $x$ increases. It meets $1$ at most once, and $x=2$ works: $\tfrac9{25}+\tfrac{16}{25}=1$ .

Answer: $x=2$

The decreasing curve y equals three fifths to the power x plus four fifths to the power x, which crosses the dashed line y equals 1 exactly once, at the marked point x equals 2

Example 53 (IITianForum). Find the number of real solutions of

\tfrac14\cdot5^{1+\cos\pi x}+2^{x^2-3}+4^{-|x|}=\tfrac34

▸ SHOW SOLUTION

Step 1. First term

$1+\cos\pi x\ge0$ , so $\tfrac14\cdot5^{1+\cos\pi x}\ge\tfrac14$ , with equality when $\cos\pi x=-1$ , that is $x$ an odd integer.

Step 2. Last two terms by AM ≥ GM

2^{x^2-3}+2^{-2|x|}\ge2\sqrt{2^{x^2-2|x|-3}}

$x^2-2|x|-3=(|x|-1)^2-4\ge-4$ , so the sum is $\ge2\cdot2^{-2}=\tfrac12$ , with equality only at $|x|=1$ .

Step 3. Combine

The left side is $\ge\tfrac14+\tfrac12=\tfrac34$ . Equality needs $|x|=1$ and $x$ odd: $x=\pm1$ .

Answer: $2$

Which method to use

A decision chart: from the question asked, arrows lead to eight situations, each paired with its method: a polynomial or quotient against zero, factorise and use the wavy curve; a square root against an expression, split on the sign of the other side; one modulus, the two standard equivalences or a substitution; several moduli, critical points and cases; a logarithm, write the domain then one base; a log inequality, split on the base; a power appearing twice, put t equal to a to the x with t positive; sum of three cubes or three squares, use the abc identity

Given	Method
polynomial against $0$	factorise, wavy curve
quotient against a number	one side, then wave on $P\cdot Q$
repeated root	even power touches, odd power crosses
$\sqrt f$ against $g$	split on the sign of $g$
$\lvert x-a\rvert^2$ and $\lvert x-a\rvert$	$t=\lvert x-a\rvert\ge0$
several moduli	critical points, one case per interval
$\lvert a\rvert+\lvert b\rvert=\lvert a+b\rvert$	$ab\ge0$
number of roots	graph, count the meets
$[x]$ in an equation	solve for $n=[x]$ first
logs with different bases	change to one base
$a^{\log_b c}$	swap to $c^{\log_b a}$
variable base	$t=\log_{u}v$ , then $\log_v u=\frac1t$
log inequality	domain, then split on the base
$a^{2x}$ and $a^x$	$t=a^x>0$
sum of powers equal to a constant	AM $\ge$ GM or a sum of squares
remainder by a quadratic	$\alpha x+\beta$ , put the two roots

What JEE asks

Across the 2025 and 2026 papers the chapter appears in five forms.

Modulus equations counted by cases or by a graph (Examples 23 to 31). This is the most frequent form in JEE Main 2026.
Logarithmic equations with a variable base, solved by $t+\frac kt$ (Examples 42 and 43).
Domains that reduce to log and rational inequalities (Examples 47 and 48).
Log laws and the swap rule inside a numerical value question (Example 39).
Polynomial division and the factor theorem (Examples 14 and 15).

The same methods appear inside domain questions in the Functions note and inside sign questions in calculus.

Quick recall

Result	Statement
$\sqrt{a+2\sqrt b}$	$\sqrt x+\sqrt y$ , $x+y=a$ , $xy=b$
$a+b+c=0$	$a^3+b^3+c^3=3abc$
$a^2+b^2+c^2-ab-bc-ca$	$\frac12\sum(a-b)^2\ge0$
componendo and dividendo	$\frac ab=\frac cd\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}$
remainder by $x-a$	$P(a)$
$a^n+b^n$ , $n$ odd	divisible by $a+b$
wavy curve	start $+$ on the right
even power	touches, no sign change
denominator roots	always hollow
$\sqrt{x^2}$	$\lvert x\rvert$
$\lvert x\rvert<a$	$-a<x<a$
$\lvert x\rvert>a$	$x<-a$ or $x>a$
$\lvert x+y\rvert=\lvert x\rvert+\lvert y\rvert$	$xy\ge0$
$\log_a N$ exists	$N>0$ , $a>0$ , $a\ne1$
$\log_a x^2$	$2\log_a\lvert x\rvert$
change of base	$\log_a N=\frac{\log_b N}{\log_b a}$
swap	$a^{\log_b c}=c^{\log_b a}$
$\log_a x>0$	$x$ and $a$ on the same side of $1$
log inequality, $0<a<1$	direction reverses
digits of $N$	characteristic $+1$
$t=a^x$	$t>0$

Before you leave this chapter

Without a pen, answer these four: the solution of $\dfrac{x-2}{x+3}\le0$ , the number of roots of $|x|^2-3|x|+2=0$ , the value of $5^{\log_{25}49}$ , and the solution of $\log_{1/3}x>2$ . If $(-3,2]$ , $4$ , $7$ and $\left(0,\tfrac19\right)$ come quickly, the chapter is yours.

Practise these methods on real exam questions in the doMath Basic Algebra and Logarithms chapter. Domains and ranges built on them are in the Functions note, and the full theory of means is in the Relation Between AM, GM and HM note.

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Basics & Logarithms Sets & Relations Theory of Equations

Basic Algebra and Logarithms for JEE: Modulus, Inequalities and Logs

Numbers and intervals

Intervals

Step 1. Part (a)

Step 2. Part (b)

Indices and surds

Step 1. Same exponent

Step 2. Use the conjugate

Step 1. First surd

Step 2. Second surd

Step 3. The sum

Step 1. Put x = −y

Step 2. Rewrite the left side

Step 3. Compare

Step 1. Substitute

Step 2. Back to x

Identities and factorisation

Ratio and proportion.** If ab=cd\dfrac ab=\dfrac cdba​=dc​, then by **componendo and dividendo

Step 1. Square and cube

Step 2. Multiply the two

Step 1. The three terms add to zero

Step 2. Solve

Step 1. Factorise

Step 2. Split

Step 3. Telescope

Step 1. Componendo and dividendo

Step 2. Square

Step 1. Substitute

Step 2. Sum of squares

Step 1. Negative x and zero

Step 2. Positive x

Step 3. Equality

Polynomials

Step 1. Two conditions

Step 2. Solve

Step 1. Form of the remainder

Step 2. Put x = 1 and x = 2

Step 1. The quotient is linear

Step 2. Compare coefficients

Step 3. Count

Step 1. Statement I

Step 2. Statement II: the conjugate

Step 3. R + R′ is an even integer

Step 4. Conclude

Step 1. One substitution

Step 2. Back to x

Step 3. Product

Step 1. Build a polynomial with known roots

Step 2. Find c

Step 3. Put x = 6

Inequalities

The wavy curve method

Step 1. Make every factor x − a

Step 2. Wave

Repeated factors and rational inequalities

Step 1. Mark the roots

Step 2. Wave

Step 3. Collect

Step 1. One side

Step 2. Wave

Inequalities with a square root

Step 1. Case x + 1 < 0

Step 2. Case x + 1 ≥ 0

Step 3. Union

Step 1. Conditions

Step 2. Square

Step 3. Intersect

Modulus

Modulus equations

Step 1. Substitute

Step 2. Roots

Step 1. Case x ≥ 1

Step 2. Case −3 ≤ x < 1

Step 3. Case x < −3

Step 1. Set B: a hidden square

Step 2. Set A

Step 3. Union

Modulus inequalities

Step 1. Set A

Step 2. Set B

Ratio and proportion. If $\dfrac ab=\dfrac cd$ , then by componendo and dividendo