Solve this logarithmic equation: ${2}^{2-\mathrm{ln}x}+{2}^{2+\mathrm{ln}x}=8$. I thought to write

$$\frac{{2}^{2}}{{2}^{\mathrm{ln}(x)}}}+{2}^{2}\cdot {2}^{\mathrm{ln}(x)}={2}^{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{{2}^{2}+{2}^{2}\cdot {\left({2}^{\mathrm{ln}(x)}\right)}^{2}}{{2}^{\mathrm{ln}(x)}}}={2}^{3$$

What should I do else?

$$\frac{{2}^{2}}{{2}^{\mathrm{ln}(x)}}}+{2}^{2}\cdot {2}^{\mathrm{ln}(x)}={2}^{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{{2}^{2}+{2}^{2}\cdot {\left({2}^{\mathrm{ln}(x)}\right)}^{2}}{{2}^{\mathrm{ln}(x)}}}={2}^{3$$

What should I do else?