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Value of a/b + b/a from a Logarithmic Mean Condition | JEE Advanced

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

Question
If ℓn(a+b3)=(ℓna+ℓnb2)\ell n\left(\dfrac{a+b}{3}\right) = \left(\dfrac{\ell na + \ell nb}{2}\right), then ab+ba\dfrac{a}{b} + \dfrac{b}{a} is equal to
Solution
Answer: 7
Step 1: Combine the right-hand side into a single logarithm.
ℓna+ℓnb2=ℓn(ab)2=ℓn(ab)1/2=ℓnab.\frac{\ell na + \ell nb}{2} = \frac{\ell n(ab)}{2} = \ell n\left(ab\right)^{1/2} = \ell n\sqrt{ab} .
 Step 2: Remove the logarithms, which is true because ℓn\ell n is one-one.
a+b3=ab.\frac{a+b}{3} = \sqrt{ab} .
 Step 3: Square both sides to clear the radical.
(a+b)29=ab ⟹ (a+b)2=9ab.\frac{(a+b)^{2}}{9} = ab \ \Longrightarrow\ (a+b)^{2} = 9ab .
 Step 4: Expand and rearrange.
a2+2ab+b2=9ab ⟹ a2+b2=7ab.a^{2}+2ab+b^{2} = 9ab \ \Longrightarrow\ a^{2}+b^{2} = 7ab .
 Step 5: Form the required expression, dividing by abab (non-zero, since both logarithms exist).
ab+ba=a2+b2ab=7abab=7.\frac{a}{b}+\frac{b}{a} = \frac{a^{2}+b^{2}}{ab} = \frac{7ab}{ab} = 7 .
 Answer: 77.
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