Basics & LogarithmsmediumJEE Main 2024Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2024)

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

Question
The sum of all the solutions of the equation (8)2x−16⋅(8)x+48=0(8)^{2x}-16\cdot(8)^{x}+48=0 is
Alog⁡86\log_{8}6
Blog⁡84\log_{8}4
C1+log⁡861+\log_{8}6correct
D1+log⁡681+\log_{6}8
Solution
Step 1: Let u=8xu=8^{x}. The equation becomes:
u2−16u+48=0u^{2}-16u+48=0
 Step 2: Factor:
(u−4)(u−12)=0  ⇒  u=4 or u=12(u-4)(u-12)=0\;\Rightarrow\;u=4\text{ or }u=12
 Step 3: Solve for xx:
8x=4  ⇒  x=log⁡848^{x}=4\;\Rightarrow\;x=\log_{8}4
8x=12  ⇒  x=log⁡8128^{x}=12\;\Rightarrow\;x=\log_{8}12
 Step 4: Sum:
x1+x2=log⁡84+log⁡812=log⁡8(4⋅12)=log⁡848x_{1}+x_{2}=\log_{8}4+\log_{8}12=\log_{8}(4\cdot 12)=\log_{8}48
 Step 5: Simplify log⁡848\log_{8}48:
log⁡848=log⁡8(8⋅6)=log⁡88+log⁡86=1+log⁡86\log_{8}48=\log_{8}(8\cdot 6)=\log_{8}8+\log_{8}6=1+\log_{8}6
 Answer: (3) 1+log⁡861+\log_{8}6
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