2d3vljtq

Answered

2022-07-11

Finding ${y}^{\mathrm{\prime}}$ of $y={\mathrm{log}}_{7}{e}^{8x}$

I know that $\frac{d}{dx}}({e}^{8x})=8{e}^{8x$, 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 $\frac{d}{dx}}(\mathrm{ln}x)={\displaystyle \frac{{y}^{\mathrm{\prime}}}{y}}={\displaystyle \frac{1}{x}}.$ (Can you show why ${\mathrm{log}}_{e}x=\mathrm{ln}x$ )

$y={\mathrm{log}}_{b}(x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{b}^{y}=x\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}({b}^{y})=\mathrm{ln}(x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y\mathrm{ln}(b)=\mathrm{ln}(x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y={\displaystyle \frac{\mathrm{ln}(x)}{\mathrm{ln}(b)}}$

$\frac{d}{dx}}{\mathrm{log}}_{b}(x)={\displaystyle \frac{1}{x\mathrm{ln}(b)}$

${\mathrm{log}}_{a}({x}^{p})=p{\mathrm{log}}_{a}x\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{7}({e}^{8x})=8x{\mathrm{log}}_{7}e$

${y}^{\mathrm{\prime}}={\displaystyle \frac{8{e}^{8x}}{{e}^{8x}\mathrm{ln}(7)}}={\displaystyle \frac{8}{\mathrm{ln}(7)}}$

I know that $\frac{d}{dx}}({e}^{8x})=8{e}^{8x$, 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 $\frac{d}{dx}}(\mathrm{ln}x)={\displaystyle \frac{{y}^{\mathrm{\prime}}}{y}}={\displaystyle \frac{1}{x}}.$ (Can you show why ${\mathrm{log}}_{e}x=\mathrm{ln}x$ )

$y={\mathrm{log}}_{b}(x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{b}^{y}=x\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}({b}^{y})=\mathrm{ln}(x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y\mathrm{ln}(b)=\mathrm{ln}(x)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y={\displaystyle \frac{\mathrm{ln}(x)}{\mathrm{ln}(b)}}$

$\frac{d}{dx}}{\mathrm{log}}_{b}(x)={\displaystyle \frac{1}{x\mathrm{ln}(b)}$

${\mathrm{log}}_{a}({x}^{p})=p{\mathrm{log}}_{a}x\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\mathrm{log}}_{7}({e}^{8x})=8x{\mathrm{log}}_{7}e$

${y}^{\mathrm{\prime}}={\displaystyle \frac{8{e}^{8x}}{{e}^{8x}\mathrm{ln}(7)}}={\displaystyle \frac{8}{\mathrm{ln}(7)}}$

Answer & Explanation

Jordan Mcpherson

Expert

2022-07-12Added 16 answers

Yes, it's fine. An easier way to proceed is by noting that

$y=x\cdot (8{\mathrm{log}}_{7}e)=x\cdot \left(8\frac{\mathrm{ln}e}{\mathrm{ln}7}\right)=x\cdot \frac{8}{\mathrm{ln}7}$

by the change of base formula for logarithms.

$y=x\cdot (8{\mathrm{log}}_{7}e)=x\cdot \left(8\frac{\mathrm{ln}e}{\mathrm{ln}7}\right)=x\cdot \frac{8}{\mathrm{ln}7}$

by the change of base formula for logarithms.

Desirae Washington

Expert

2022-07-13Added 5 answers

Simply we have

$y={\mathrm{log}}_{7}{e}^{8x}=\frac{\mathrm{ln}{e}^{8x}}{\mathrm{ln}7}=\frac{8x}{\mathrm{ln}7}\Rightarrow {y}^{\prime}=\frac{8}{\mathrm{ln}7}$

$y={\mathrm{log}}_{7}{e}^{8x}=\frac{\mathrm{ln}{e}^{8x}}{\mathrm{ln}7}=\frac{8x}{\mathrm{ln}7}\Rightarrow {y}^{\prime}=\frac{8}{\mathrm{ln}7}$

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