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Prove If $m$ is an integer with $m\neq0$,then $E(1/m)=(E(1))^{1/m}$.

When $m$ is positive I try by using property $E(mx)=(E(x))^m$ but this property works just for positive $m$, so I am stuck here because $1/m$ is not an integer when I let $x=1 $.

When $m$ is negative I try using property $E(-x)=1/E(x)$ because this property works for integers both positive and negative.

But how can I go from $(E(1/m))^{-1}$ to $(E(1))^{1/m}$

any hint please.

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    Please use appropriate formatting. Otherwise it is not clear what you are asking. – 2017-02-13
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    I updated the formatting for you, and it will be visible once approved. Please study it so that you can use it in the future. Best. – 2017-02-13
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    What is $E?$ This question needs restating pronto. – 2017-02-14
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    My guess is that $E(x)$ means $e^x$ or $\exp(x)$, the exponential function base $e$. Using $E(x)$ is sufficiently unconventional to require an explanation within the body of the Question, unless the intention is to leave the Readers puzzled. – 2017-02-14
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    How can I use it? – 2017-02-14

1 Answers 1

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I assume $E$ stands for the exponential function? If so, then if $m$ is a positive integer we have

$E(1/m)^m = E(1/m) \cdots E(1/m) = E(1/m + \cdots + 1/m) = E(m\cdot(1/m)) = E(1)$,

so that $E(1/m) = (E(1/m)^m)^{\frac{1}{m}}=E(1)^{\frac{1}{m}}$.

Knowing the desired relation is true for positive integers, if $m$ is negative then $-m$ is positive, and

$E\left(\frac{1}{m}\right)^m = E\left(-(\frac{1}{-m})\right)^m = \left(\left(E\left(\frac{1}{-m}\right)\right)^{-1}\right)^m = E\left(\frac{1}{-m}\right)^{-m} = E(1)$, and again we have

$E\left(\frac{1}{m}\right) = \left(E\left(\frac{1}{m}\right)^m\right)^{1/m} = E(1)^{\frac{1}{m}}$.