# Solve this logarithmic equation: 2^(2- ln x)+2^(2+ ln x)=8

Solve this logarithmic equation: ${2}^{2-\mathrm{ln}x}+{2}^{2+\mathrm{ln}x}=8$. I thought to write
$\frac{{2}^{2}}{{2}^{\mathrm{ln}\left(x\right)}}+{2}^{2}\cdot {2}^{\mathrm{ln}\left(x\right)}={2}^{3}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{{2}^{2}+{2}^{2}\cdot {\left({2}^{\mathrm{ln}\left(x\right)}\right)}^{2}}{{2}^{\mathrm{ln}\left(x\right)}}={2}^{3}$
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wespee0
We have ${2}^{2-\mathrm{ln}\left(x\right)}=\frac{4}{{2}^{\mathrm{ln}\left(x\right)}}$ and ${2}^{2+\mathrm{ln}\left(x\right)}=4\cdot {2}^{\mathrm{ln}\left(x\right)}$. Hence, setting ${2}^{\mathrm{ln}\left(x\right)}=a$, we obtain
$\frac{4}{a}+4a=8\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{a}^{2}+1=2a\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}a=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{2}^{\mathrm{ln}\left(x\right)}=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=1$
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Maribel Vang
Hint:
Dividing by 4 we get the equivalent equation
${2}^{-\mathrm{ln}x}+{2}^{\mathrm{ln}x}=2$
then let $y={2}^{\mathrm{ln}x}$ so
${y}^{-1}+y=2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}1+{y}^{2}=2y$