Question

e^xloga+e^alogx+e^aloga

Answer 1

$Answer \: \: \\ \\ e {}^{x log_{e}(a) } + e {}^{a log_{e}(x) } + e {}^{a log_{e}(a) } \\ \\ e {}^{ log_{e}(a {}^{x} ) } + e {}^{ log_{e}(x {}^{a} ) } + e {}^{ log_{e}(a {}^{a} ) } \\ \\ a {}^{x} + x {}^{a} + a {}^{a} \\ \\ NOTE \: \\ \\ 1) \: \: \: n log(x) = log(x {}^{n} ) \\ \\ 2) \: \: \: e {}^{ log_{e}(y) } = y$