How do you solve ${(e^{3})}^{2 x} = (e^{3}) (e^{2 x})$ $(e^3)^(2x) = (e^3)(e^(2x))$ ?

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Answer 1 · 2016-06-08T14:02:27

$x = \frac{3}{4}$ $x = 3/4$

${(e^{3})}^{2 x} = e^{3 \times 2 x} = e^{6 x}$ $(e^3)^{2x} = e^{3 xx 2x} = e^{6x}$

so

$e^{6 x} = e^{4 x} e^{2 x} = (e^{3}) (e^{2 x})$ $e^{6x}= e^{4x}e^{2x} = (e^3)(e^{2x})$

then

$(e^{4 x} - e^{3}) e^{2 x} = 0$ $(e^{4x}-e^3)e^{2x} =0$

but

$e^{2 x} > 0$ $e^{2x} > 0$

so

$e^{4 x} = e^{3} \to 4 x = 3 \to x = \frac{3}{4}$ $e^{4x}=e^3->4x=3->x = 3/4$

How do you solve (e3)2x=(e3)(e2x)(e^3)^(2x) = (e^3)(e^(2x))?