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Chained Logarithmic Conditions Evaluating a Power | JEE Advanced Logarithms

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

Question
If log⁡α8=γ\log_{\alpha}8 = \gamma, log⁡βα=−1\log_{\beta}\alpha = -1 and log⁡1/4β=−1\log_{1/4}\beta = -1 then
(1α+1)log⁡5(β2+4γ2)\left(\frac{1}{\alpha}+1\right)^{\log_{\sqrt5}\left(\beta^{2}+4\gamma^{2}\right)}
 is equal to
A5\sqrt5
B55
C2525
D625625correct
Solution
Step 1: Unwind the conditions in the order that each one is solvable. Start with the last, which involves only β\beta:
log⁡1/4β=−1 ⟹ β=(14)−1=4.\log_{1/4}\beta = -1 \ \Longrightarrow\ \beta = \left(\frac14\right)^{-1} = 4 .
 Step 2: Use β\beta to get α\alpha.
log⁡βα=−1 ⟹ α=β−1=14.\log_{\beta}\alpha = -1 \ \Longrightarrow\ \alpha = \beta^{-1} = \frac14 .
 Step 3: Use α\alpha to get γ\gamma.
γ=log⁡α8=log⁡1/48.\gamma = \log_{\alpha}8 = \log_{1/4}8 .
 Writing both sides in base 22: (2−2)γ=23\left(2^{-2}\right)^{\gamma} = 2^{3}, so −2γ=3-2\gamma = 3 and
γ=−32.\gamma = -\frac32 .
 Step 4: Evaluate the base of the power.
1α+1=4+1=5.\frac{1}{\alpha}+1 = 4+1 = 5 .
 Step 5: Evaluate the quantity inside the logarithm.
β2+4γ2=16+4(−32)2=16+4⋅94=16+9=25.\beta^{2}+4\gamma^{2} = 16 + 4\left(-\frac32\right)^{2} = 16 + 4\cdot\frac94 = 16+9 = 25 .
 Step 6: Evaluate the exponent, remembering the base is 5\sqrt5.
log⁡525=log⁡525log⁡55=212=4.\log_{\sqrt5}25 = \frac{\log_5 25}{\log_5\sqrt5} = \frac{2}{\tfrac12} = 4 .
 Step 7: Combine.
54=625.5^{4} = 625 .
 (Reading the base as 55 rather than 5\sqrt5 gives an exponent of 22 and the answer 2525, which is option (3) - the intended trap.) Answer: (4).
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