Raina Gomez

2022-09-20

Solving ${e}^{4x}+3{e}^{2x}-28=0$
How to solve this equation:
${e}^{4x}+3{e}^{2x}-28=0$
I don't know how to solve this problem. I read over another example, ${e}^{2x}-2{e}^{x}-8=0,$ and it said that ${e}^{2x}$ is $e$ to the $x$ squared, so ${e}^{2x}={e}^{{x}^{2}}$. Why is this so? So ${e}^{4x}={e}^{{x}^{4}}$?

Wischarm1q

Hint:
$\begin{array}{rl}{e}^{4x}+3{e}^{2x}-28=0& \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left({e}^{2x}{\right)}^{2}+3\left({e}^{2x}\right)-28=0\\ & \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{e}^{2x}=\dots \\ & \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}=\dots \end{array}$
And I think what "$e$ to the $x$ squared" means is $\left({e}^{x}{\right)}^{2}$, in this case it is equal to ${e}^{2x}$ since one have $\left({a}^{m}{\right)}^{n}={a}^{mn}.$

saucletbh

$\left({e}^{2x}{\right)}^{2}+3{e}^{2x}-28=0$
${u}^{2}+3u-28=0$
This is a quadratic equation. Once you've found $u$, you've got ${e}^{2x}$. After that, find $x$.

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