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 sum of the roots of the equation x+1−2log⁡2(3+2x)+2log⁡4(10−2−x)=0x+1-2\log_{2}(3+2^{x})+2\log_{4}(10-2^{-x})=0 is
Alog⁡214\log_{2}14
Blog⁡211\log_{2}11correct
Clog⁡212\log_{2}12
Dlog⁡213\log_{2}13
Solution
Step 1: Rewrite using log⁡4u=12log⁡2u\log_{4}u=\dfrac{1}{2}\log_{2}u:
x+1−2log⁡2(3+2x)+log⁡2(10−2−x)=0x+1-2\log_{2}(3+2^{x})+\log_{2}(10-2^{-x})=0
 Step 2: Let 2x=t2^{x}=t, so 2−x=1t2^{-x}=\dfrac{1}{t}:
log⁡2t+1−2log⁡2(3+t)+log⁡2(10−1t)=0\log_{2}t+1-2\log_{2}(3+t)+\log_{2}\left(10-\dfrac{1}{t}\right)=0
 Step 3: Combine logs:
log⁡2[2t⋅(10−1t)(3+t)2]=0\log_{2}\left[\dfrac{2t\cdot\left(10-\dfrac{1}{t}\right)}{(3+t)^{2}}\right]=0
2t(10−1t)(3+t)2=1  ⇒  20t−2(3+t)2=1\dfrac{2t\left(10-\dfrac{1}{t}\right)}{(3+t)^{2}}=1\;\Rightarrow\;\dfrac{20t-2}{(3+t)^{2}}=1
 Step 4: Expand:
20t−2=(3+t)2=9+6t+t220t-2=(3+t)^{2}=9+6t+t^{2}
t2−14t+11=0t^{2}-14t+11=0
 Step 5: Roots are t1,t2t_{1},t_{2} with t1t2=11t_{1}t_{2}=11 and t1+t2=14t_{1}+t_{2}=14. Since t=2xt=2^{x}: x1+x2=log⁡2t1+log⁡2t2=log⁡2(t1t2)=log⁡211x_{1}+x_{2}=\log_{2}t_{1}+\log_{2}t_{2}=\log_{2}(t_{1}t_{2})=\log_{2}11. Answer: (2) log⁡211\log_{2}11
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