2d3vljtq

2022-07-11

Finding ${y}^{\mathrm{\prime }}$ of $y={\mathrm{log}}_{7}{e}^{8x}$
I know that $\frac{d}{dx}\left({e}^{8x}\right)=8{e}^{8x}$, but I am confused on how to work the rest of the problem.
Is this correct:
${\mathrm{log}}_{e}x=\mathrm{ln}x$ and that $\frac{d}{dx}\left(\mathrm{ln}x\right)=\frac{{y}^{\mathrm{\prime }}}{y}=\frac{1}{x}.$ (Can you show why ${\mathrm{log}}_{e}x=\mathrm{ln}x$ )
$y={\mathrm{log}}_{b}\left(x\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{b}^{y}=x\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}\left({b}^{y}\right)=\mathrm{ln}\left(x\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}y\mathrm{ln}\left(b\right)=\mathrm{ln}\left(x\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}y=\frac{\mathrm{ln}\left(x\right)}{\mathrm{ln}\left(b\right)}$
$\frac{d}{dx}{\mathrm{log}}_{b}\left(x\right)=\frac{1}{x\mathrm{ln}\left(b\right)}$
${\mathrm{log}}_{a}\left({x}^{p}\right)=p{\mathrm{log}}_{a}x\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{7}\left({e}^{8x}\right)=8x{\mathrm{log}}_{7}e$
${y}^{\mathrm{\prime }}=\frac{8{e}^{8x}}{{e}^{8x}\mathrm{ln}\left(7\right)}=\frac{8}{\mathrm{ln}\left(7\right)}$

Jordan Mcpherson

Expert

Yes, it's fine. An easier way to proceed is by noting that
$y=x\cdot \left(8{\mathrm{log}}_{7}e\right)=x\cdot \left(8\frac{\mathrm{ln}e}{\mathrm{ln}7}\right)=x\cdot \frac{8}{\mathrm{ln}7}$
by the change of base formula for logarithms.

Desirae Washington

Expert

Simply we have
$y={\mathrm{log}}_{7}{e}^{8x}=\frac{\mathrm{ln}{e}^{8x}}{\mathrm{ln}7}=\frac{8x}{\mathrm{ln}7}⇒{y}^{\prime }=\frac{8}{\mathrm{ln}7}$

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