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Question : 43 of 86

Marks: +1, -0

If

a^{b}=b^{a}+

then the value of

\left(\frac{a}{b}\right)^{\frac{a}{b}}-a^{\frac{a}{b}-1}

$-1$
0
1
None.

Solution:

Concept:

Use the relation

a^b = b^a

to simplify the expression by rewriting powers.

Explanation:

Given

a^b = b^a

.

We need to find the value of

E = \left(\frac{a}{b}\right)^{\frac{a}{b}} - a^{\frac{a}{b}-1}

.

Rewrite the first term:

\left(\frac{a}{b}\right)^{\frac{a}{b}} = \frac{a^{\frac{a}{b}}}{b^{\frac{a}{b}}}

.

Rewrite the second term:

a^{\frac{a}{b}-1} = \frac{a^{\frac{a}{b}}}{a}

.

So

E = a^{\frac{a}{b}} \left( \frac{1}{b^{\frac{a}{b}}} - \frac{1}{a} \right)

.

Now from

a^b = b^a

, raise both sides to the power

\frac{1}{b}

:

a = b^{\frac{a}{b}}

.

Thus

b^{\frac{a}{b}} = a

, so

\frac{1}{b^{\frac{a}{b}}} = \frac{1}{a}

.

Therefore the bracket becomes

\frac{1}{a} - \frac{1}{a} = 0

, so

E = 0

.

Answer:

0

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