Basics & LogarithmseasyFree

Basics & Logarithms: Value Equal

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

Question

The value of

a^{\sqrt{\log_a b}} - b^{\sqrt{\log_b a}}

(where

a, b > 0

and

a \neq 1

) is equal to

Solution

Answer: 0

The value of

a^{\sqrt{\log_a b}} - b^{\sqrt{\log_b a}}

(where

a, b > 0

and

a \neq 1

) is equal to ___. Step 1: Let

y = a^{\sqrt{\log_a b}}

. Take

\log_b

of both sides:

\log_b y = \sqrt{\log_a b} \cdot \log_b a = \sqrt{\log_a b} \cdot \frac{1}{\log_a b} = \frac{1}{\sqrt{\log_a b}}

Step 2: Since

\log_b a = \dfrac{1}{\log_a b}

, we have

\dfrac{1}{\sqrt{\log_a b}} = \sqrt{\dfrac{1}{\log_a b}} = \sqrt{\log_b a}

. Therefore

\log_b y = \sqrt{\log_b a}

, which gives

y = b^{\sqrt{\log_b a}}

. Step 3: Thus

a^{\sqrt{\log_a b}} = b^{\sqrt{\log_b a}}

, so the expression equals

0

. Answer: 0

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