<PRE><B>Question 748186</B>
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4^x\ =\ 2^{2x}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^x\ =\ 16\,\cdot\,4^x]


is equivalent to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16\,\cdot\,2^x\ =\ 1]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^x\ =\ \frac{1}{16}]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -4]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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