TS ICET 23 May 2018 Shift 2 Paper

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Question : 76 of 200

Marks: +1, -0

(a^m)^\n = {a^m}^n

⇒ m =

$(\n-1)^\n$
$n^{\n —1$
$n/{\n-1}$
$1/{\n.(\n-1)}$

Solution:

Given

(a^m)^n=a^{m^n}

⇒ a^{m n}=a^{m^{n}} ⇒ mn = m^{n}⇒n=m ^{(n-1)}

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