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How would I rewrite this logarithmic equation: $\ln(37)= 3.6109$, in exponential form?

-Thanks

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    @Jasmine: When you use approximate values you need to worry about propagation of errors, which mathematicians rarely do (because their numbers don't have errors). Subtracting nearly equal numbers is the poster child. If you subtract 1-0.999 with each number accurate to a part in a million, the difference is accurate to 2 parts in a thousand.2011-05-31

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The definition of $\ln(x)$ is that it is the number $y$ such that $e^y=x$. In other words, $e^{\ln(x)}=x.$ We have the equation $\ln(37)=3.6109.$ Because both sides are equal, we have that $e^{\ln(37)}=e^{3.6109}.$ By the definition of $\ln$, this simplifies to $37=e^{3.6109}.$

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Another way to see it (that is equivalent to Zev's answer) is that
$\log_{b}(a) = x$
is equivalent to
$a = b^x$.

$\ln$ is just $\log_{e}$, so
$\ln(37) = 3.6109$
is simply $\log_b(a) = x$ with $b = e$, $a = 37$, $x = 3.6109$
and can be rewritten as
$37 = e^{3.6109}$.

That good enough for your needs?

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    It is perfect! Thnx a bunch :)2011-05-31