# How to solve for k when the area about the x axis and under the graph of the f(x)=1/x from interval x=[2,k] is equal to ln(4)?

How to solve for $k$ when the area about the $x$ axis and under the graph of the $f\left(x\right)=\frac{1}{x}$ from interval $x=\left[2,k\right]$ is equal to $\mathrm{ln}\left(4\right)$?
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Kaylee Evans
Notice
$\underset{2}{\overset{k}{\int }}\frac{1}{x}dx=\mathrm{ln}4=2\mathrm{ln}2$
This is given. But
$\underset{2}{\overset{k}{\int }}\frac{1}{x}dx=\mathrm{ln}k-\mathrm{ln}2$
Hence,
$\mathrm{ln}k-\mathrm{ln}2=2\mathrm{ln}2\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}k=3\mathrm{ln}2\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}k=\mathrm{ln}{2}^{3}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}k={2}^{3}=8$