Let $f (x) = e^{3 x}, g (x) = {log}_{e} x, x > 0$ , then $f o g (x)$ is

3 x

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x^{3}

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{log}_{10} 3 x

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log 3 x

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Solution

The correct option is C $x^{3}$
Given $f (x) = e^{3 x}, g (x) = {log}_{e} x, x > 0$

$f o g (x) = f ({log}_{e} x) = e^{3 {log}_{e} x} = x^{3}$

Hence, option B.

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