Basics & LogarithmsmediumJEE Main 2021Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2021)

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

Question

The number of solutions of the equation

\log_{(x+1)}(2x^{2}+7x+5)+\log_{(2x+5)}(x+1)^{2}-4=0

x>0

Solution

Answer: 1

Step 1: Note

2x^{2}+7x+5=(2x+5)(x+1)

. Step 2: So

\log_{(x+1)}(2x^{2}+7x+5)=\log_{(x+1)}[(2x+5)(x+1)]=\log_{(x+1)}(2x+5)+1

. Step 3: The equation becomes:

\log_{(x+1)}(2x+5)+1+2\log_{(2x+5)}(x+1)-4=0

\log_{(x+1)}(2x+5)+2\log_{(2x+5)}(x+1)-3=0

Step 4: Let

t=\log_{(x+1)}(2x+5)

. Then

\log_{(2x+5)}(x+1)=\dfrac{1}{t}

t+\dfrac{2}{t}-3=0\;\Rightarrow\;t^{2}-3t+2=0\;\Rightarrow\;(t-1)(t-2)=0

t=1

t=2

. Step 5: Case

t=1

\log_{(x+1)}(2x+5)=1\;\Rightarrow\;2x+5=x+1\;\Rightarrow\;x=-4

(rejected, since

x>0

). Step 6: Case

t=2

\log_{(x+1)}(2x+5)=2\;\Rightarrow\;2x+5=(x+1)^{2}=x^{2}+2x+1\;\Rightarrow\;x^{2}=4\;\Rightarrow\;x=2

(accepted,

x>0

). Step 7: Only

1

solution. Answer:

1

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