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


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Can you elaborate as to what you mean by "logarithmic equation"? The way I see it, it already IS one... – El'endia Starman May 31 '11 at 0:28
It looks logarithmic to me already. Can you give an example of what a logarithmic equation looks like? Yelling help is not productive. – Ross Millikan May 31 '11 at 0:29
Oops, sorry 'bout that. Thanks for pointing that out. – Jasmine V. May 31 '11 at 0:44
Note that the equation, as given, is not correct. It is close, but not exact. – Ross Millikan May 31 '11 at 1:07
@Jasmine: The actual value of $\ln(37)$ is a real number whose decimal expansion is infinite; it begins $$\ln(37)=3.6109179126442244443680956710314\ldots$$ So the value 3.6109 is only an approximation to the correct value. – Zev Chonoles May 31 '11 at 3:10

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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Thanks soo much! – Jasmine V. May 31 '11 at 2:51

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 :) – Jasmine V. May 31 '11 at 2:51

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