Basics & LogarithmshardFree

Basics & Logarithms: Value

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

Question
If A=(log⁡31−log⁡34)(log⁡39−log⁡32)(log⁡31−log⁡39)(log⁡38−log⁡34)A = \dfrac{(\log_3 1 - \log_3 4)(\log_3 9 - \log_3 2)}{(\log_3 1 - \log_3 9)(\log_3 8 - \log_3 4)}, then the value of 2(3A)2(3^A) is
Solution
Answer: 9
Step 1: Evaluate each logarithm. log⁡31=0\log_3 1 = 0, log⁡34=2log⁡32\log_3 4 = 2\log_3 2, log⁡39=2\log_3 9 = 2, log⁡38=3log⁡32\log_3 8 = 3\log_3 2.
A=(0−2log⁡32)(2−log⁡32)(0−2)(3log⁡32−2log⁡32)=−2log⁡32⋅log⁡3(9/2)−2log⁡32A = \frac{(0 - 2\log_3 2)(2 - \log_3 2)}{(0 - 2)(3\log_3 2 - 2\log_3 2)} = \frac{-2\log_3 2 \cdot \log_3(9/2)}{-2\log_3 2}
 Step 2: Simplify (cancelling −2log⁡32-2\log_3 2 from numerator and denominator):
A=log⁡3 ⁣(92)A = \log_3\!\left(\frac{9}{2}\right)
 Step 3: Therefore 3A=923^A = \dfrac{9}{2}, and:
2(3A)=2⋅92=92(3^A) = 2 \cdot \frac{9}{2} = 9
 Answer: 9
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