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I know that the $\ln{(7x)}$ will go to just $7x$, but I'm confused about how the $17$ fits into the answer.

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    $\begingroup$ $x^{a+b}=x^ax^b$ $\endgroup$ Aug 21, 2018 at 15:38
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    $\begingroup$ You didn't say what the problem statement was. How can anyone help you if they don't know what exactly it is that you're trying to do? $\endgroup$ Aug 21, 2018 at 15:41

2 Answers 2

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$$e^{17+\ln 7x} = e^{17} e^{ \ln 7x} = 7xe^{17}$$

And it does not simplify any more.

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I assume you mean $e^{17+\ln(7x)}$ and you want to simplify?

You can use the following:

$e^{a+b} = e^ae^b$

$e^{\ln(x)}=x$

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