Basics & LogarithmsmediumFree

Basics & Logarithms — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Sum of all the roots of the equation log⁡7(2x−1)+log⁡7(2x−7)=1\log_7(2^x-1) + \log_7(2^x-7) = 1 is:
A88
B33correct
C44
D11
Solution
Step 1: Combine the logarithms
log⁡7[(2x−1)(2x−7)]=1  ⟹  (2x−1)(2x−7)=7\log_7\big[(2^x-1)(2^x-7)\big] = 1 \implies (2^x-1)(2^x-7) = 7
 Step 2: Substitute t=2xt = 2^x
t2−8t+7=7  ⟹  t(t−8)=0  ⟹  t=8(t=0 rejected)t^2-8t+7 = 7 \implies t(t-8) = 0 \implies t = 8 \quad (t = 0 \text{ rejected})
2x=8  ⟹  x=32^x = 8 \implies x = 3
 This satisfies the domain (2x>72^x > 7), so the sum of all roots is 3. Answer: (2)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Basics & Logarithms · easy
The product of all solutions of the equation e 5( e x) 2 +3 =x 8 , x>0 , is
Basics & Logarithms · easy
The number of real solutions of the equation x (x 2 +3|x|+5|x-1|+6|x-2| )=0 is
Basics & Logarithms · easy
If a+b+c=1 , ab+bc+ca=2 and abc=3 , then the value of a 4 +b 4 +c 4 is equal to
Basics & Logarithms · easy
Sum of all the values of x satisfying the equation 17 11 ( x+11 + x ) = 0 is:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath