How do you solve $e^{x^{2}} = \frac{e^{15}}{{(e^{x})}^{2}}$ $e^ { x ^ { 2} } = \frac { e ^ { 15} } { ( e ^ { x } ) ^ { 2} }$ ?

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Answer 1 · 2017-12-09T00:58:37

$x = 3 or x = - 5$ $x=3 or x=-5$

$e^{x^{2}} = \frac{e^{15}}{e^{2 x}}$ $e^(x^2)=e^15/e^(2x)$ $\Leftrightarrow$ $<=>$

$\Leftrightarrow$ $<=>$ ${ln e}^{x^{2}} = ln (\frac{e^{15}}{e^{2 x}})$ $lne^(x^2)=ln(e^15/e^(2x))$ $\Leftrightarrow$ $<=>$

$\Leftrightarrow$ $<=>$ $x^{2} = {ln e}^{15} - {ln e}^{2 x}$ $x^2=lne^15-lne^(2x)$ $\Leftrightarrow$ $<=>$

$\Leftrightarrow$ $<=>$ $x^{2} = 15 ln e - 2 x \cdot ln e$ $x^2=15lne-2x*lne$ $\Leftrightarrow$ $<=>$

$\Leftrightarrow$ $<=>$ $x^{2} = 15 - 2 x$ $x^2 = 15-2x$ $\Leftrightarrow$ $<=>$

$\Leftrightarrow$ $<=>$ $x^{2} + 2 x - 15 = 0$ $x^2+2x-15=0$ $\Leftrightarrow$ $<=>$

$\Leftrightarrow$ $<=>$ $(x - 3) (x + 5) = 0$ $(x-3)(x+5)=0$ $\Leftrightarrow$ $<=>$

$\Leftrightarrow$ $<=>$ $x = 3$ $x=3$ or $x = - 5$ $x=-5$

Used logarithm properties :
$ln e = 1$ $lne=1$
${ln e}^{x} = x ln e$ $lne^x=xlne$
$ln x = y \Leftrightarrow x = e^{y}$ $lnx=y <=> x=e^y$
$ln (\frac{a}{b}) = ln a - ln b$ $ln(a/b)=lna-lnb$

How do you solve e^ { x ^ { 2} } = \frac { e ^ { 15} } { ( e ^ { x } ) ^ { 2} }ex2=e15(ex)2e^ { x ^ { 2} } = \frac { e ^ { 15} } { ( e ^ { x } ) ^ { 2} }?