Basics & LogarithmsmediumPYQ · JEE Main · 20 Jul 2021 · Shift 2 (Afternoon)Free

Basics & Logarithms: Number Solutions Equation (JEE Main 2021)

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

Question
The number of solutions of the equation log⁡(x+1)(2x2+7x+5)+log⁡(2x+5)(x+1)2−4=0\log_{(x+1)}(2x^{2}+7x+5)+\log_{(2x+5)}(x+1)^{2}-4=0, x>0x>0 is
Solution
Answer: 1
Step 1: Note 2x2+7x+5=(2x+5)(x+1)2x^{2}+7x+5=(2x+5)(x+1). Step 2: So log⁡(x+1)(2x2+7x+5)=log⁡(x+1)[(2x+5)(x+1)]=log⁡(x+1)(2x+5)+1\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+2log⁡(2x+5)(x+1)−4=0\log_{(x+1)}(2x+5)+1+2\log_{(2x+5)}(x+1)-4=0
log⁡(x+1)(2x+5)+2log⁡(2x+5)(x+1)−3=0\log_{(x+1)}(2x+5)+2\log_{(2x+5)}(x+1)-3=0
 Step 4: Let t=log⁡(x+1)(2x+5)t=\log_{(x+1)}(2x+5). Then log⁡(2x+5)(x+1)=1t\log_{(2x+5)}(x+1)=\dfrac{1}{t}:
t+2t−3=0  ⇒  t2−3t+2=0  ⇒  (t−1)(t−2)=0t+\dfrac{2}{t}-3=0\;\Rightarrow\;t^{2}-3t+2=0\;\Rightarrow\;(t-1)(t-2)=0
 So t=1t=1 or t=2t=2. Step 5: Case t=1t=1: log⁡(x+1)(2x+5)=1  ⇒  2x+5=x+1  ⇒  x=−4\log_{(x+1)}(2x+5)=1\;\Rightarrow\;2x+5=x+1\;\Rightarrow\;x=-4 (rejected, since x>0x>0). Step 6: Case t=2t=2: log⁡(x+1)(2x+5)=2  ⇒  2x+5=(x+1)2=x2+2x+1  ⇒  x2=4  ⇒  x=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>0x>0). Step 7: Only 11 solution. Answer: 11
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