$e^{x log a} + e^{a log x} + e^{a log a}$ = ?

a^{x} + x^{a} + a^{a}

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a^{x} + x^{a} + 2 a^{a}

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a^{2} x + x^{a} + a^{a}

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x^{a} + a^{a}

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Solution

The correct option is A $a^{x} + x^{a} + a^{a}$
$e^{x log a} + e^{a log x} + e^{a log a}$
$= e^{log a^{x}} + e^{log x^{a}} + e^{log a^{a}}$ $(n log x = log x^{n})$
$= a^{x} + x^{a} + a^{a}$ $(a^{{log}_{a} x} = x)$

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