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$$ \ln(x) = 1 + \ln(5)$$

$$ x = e^{1+ \ln(5)} = e^{1+5} = e^6$$

What exactly am I doing wrong here?

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it is $e\cdot 5$ not $e^{1+5}$ –  user31280 Jan 21 '13 at 20:34

3 Answers 3

up vote 2 down vote accepted

You change $\ln(5)$ to $5$ in the exponent, when it should be $$e^{1+\ln(5)}=e\cdot e^{\ln(5)}=5e.$$

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Thank you, can't believe I missed that. –  ZafarS Jan 21 '13 at 20:36
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@ZafarS: You're welcome; we all make mistakes at some point. –  Clayton Jan 21 '13 at 20:37

you can compute in that way $ln(x) = 1 + ln(5)$ then $ln(x) = ln(e) + ln(5)$ ln(X)=ln(5e) then x=5e

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$$ \ln x = 1 + \ln 5$$ $$ \ln x = \ln e + \ln 5$$ $$\ln x=\ln(5e)$$ $$x=5e$$

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