# Why do I get different results using the e function for the ln?

I have a question which Kind of makes me crazy right now. If I have the equation $1=\ln(2•3)$, I could then use the e function to remove the $\ln.$ Doing that I get $e^1 = 6.$ Now say we are using the rules of the log at the beginning we could write $$1 = \ln(2)+\ln(3).$$ Using the $e$ function I get $e^1 = 2+3$. How is this possible? What is my mistake?

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## 2 Answers

Your mistake is that $e^{\ln 2+\ln 3} = e^{\ln2} \cdot e^{\ln3}$ and not $e^{\ln 2+\ln 3} = e^{\ln2} + e^{\ln3}$.

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You can't have the equation like $1 = \log(2\cdot 3)$. It's like saying if you have $2= 3$ and if I add $1$ on both sides then to get $3=4$. Now can I ask how $3=4$?

Moreover if $1 = \log{2} + \log{3}$, then $e^{1} = e^{\log{2}+\log{3}} = e^{\log{2}} \cdot e^{\log{3}} = 2 \cdot 3=6$

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That's not the major mistake. –  Hendrik Vogt Jul 3 '11 at 13:02
@Hendrik: Thanks. Yes, I think, I have done it correctly now. –  user9413 Jul 3 '11 at 13:04