How do $\ln$ and $e$ cancel out when there is an exponent that is negative?

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I have a question about $\ln$ and $e$.

Essentially I know that $e$ and $\ln$ cancel/reverse each other to give $x$. Ex: $e^{\ln x} = x$

But how does this work: $-e^{-\ln 2} = -1/2$

I thought it would just give me $-(-2)$.

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2 Answers

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Recall that that $a^{-b}= (a^b)^{-1}$.

Thus $e^{-\ln 2}= (e^{\ln 2})^{-1}= 2^{-1}= 1/2$.

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We know that $e^{\ln x} = x$ and $p \ln x = ln x^p$.

Using this we get

$$ -e^{-\ln 2} = -e^{-1\ln 2}= -e^{\ln 2^{-1}} = -e^{ln(\frac{1}{2})}= -\frac{1}{2}. $$

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