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Find the transformation that takes $y=3^x$ to $y=\textit{e}^x$. I have tried:

Let $y=3^x$ to $y=e^{x'}$

$$\log_{3}(y)=x\quad\text{hence}\quad\log_{3}(y)=\frac{\log_{e}(y)}{\log_{e}(3)}$$

$$x\log_{e}(3)=x'$$

Gives the transformation as dilate by $\log_e(3)$ And also this:

$$3^{\log_{3}e}=e$$

$$e^x=3^{(\log_{3}e)x}$$

And the transformation is dilate by $\frac{1}{\log_3e}$

Could I please get an explanation which is right?

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    $\begingroup$ NOTE: $\log_3 e=\frac{1}{\log_e 3}$ $\endgroup$
    – Mufasa
    Jan 3, 2016 at 14:10
  • $\begingroup$ $y=3^x \implies \log(y)=x\log(3) \implies y=e^{x\log(3)}$ could help $\endgroup$ Jan 3, 2016 at 14:12

3 Answers 3

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Both of your answers are correct as I pointed out in my comment:$$\log_3 e=\frac{\log_e e}{\log_e 3}=\frac{1}{\log_e 3}$$

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  • $\begingroup$ Thank you, I didn't see you comment earlier. Sorry $\endgroup$ Jan 3, 2016 at 14:26
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It can be re-written as follows $$y=3^x=e^{\ln(3^x)}=e^{x\ln 3}=(e^x)^{\ln 3}$$

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Notice that $b = \exp(\ln(b))$ for all $b > 0$. From this follows that $3^x$ can be re-written as $$ 3^x = \exp( \ln(3^x) ) = \exp( x \cdot \ln(3) ), $$ where the last equality uses the logarithmic identity $\ln(b^a) = a \cdot \ln(b)$ for all $a$ and all $b > 0$.

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  • $\begingroup$ Sorry but could you please clarify your answer a bit more? I don't understand it fully sorry $\endgroup$ Jan 3, 2016 at 14:21

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