Why is $\frac{13}{10} = e^{\ln\frac{10}{13}*x}$ Why is:
$$\frac{13}{10}^{-x} = e^{ln\frac{10}{13}*x}$ $
I thought that the $\ln$ and the $e$ canceled out making it equivalent to saying:
$$e^{\ln\frac{10}{13}*x} = \frac{10x}{13}$$
Is this not correct?
 A: $$
e^{x\ln(10/13)}
=e^{\ln((10/13)^{x})}
=\left(\frac{10}{13}\right)^{x}
=\left(\frac{13}{10}\right)^{-x}
$$
A: Your question is missing the $-x$ term in the power.
Further, $\ln$ never canceled out with $e$, it is by the property: $e^{\ln x}=x$
Now in your case, $\frac{13}{10}^{-x}=\left(\left(\frac{13}{10}\right)^{-1}\right)^x=\left(\frac{10}{13}\right)^x=e^{x\ln (\frac{10}{13})}$
A: Notice that the fractions in each case are flipped, and that taking the reciprocal of a fraction is the same as raising it to the power of $-1$. Then we have
$$e^{\ln(10/13) \cdot x} = \left(e^{\ln(10/13)} \right)^x= \left( \frac{10}{13} \right)^x =  \left(  \left( \frac{13}{10} \right)^{-1} \right)^x = \left( \frac{13}{10} \right)^{-x}$$
A: You need some parentheses to show the order of operations.  You title does not match the first line of the question, but the question looks to match what is going on here.  When you write $\frac {13}{10}^{-x}$ do you mean $\left(\frac {13}{10}\right)^{-x}$ or $\frac {(13^{-x})}{10}$?  Similarly, when you write $ e^{ln\frac{10}{13}*x}$ do you mean $ e^{(\ln\frac{10}{13})*x}$ or $ e^{\ln(\frac{10}{13}*x)}$?  
For the question as edited (is that what you meant?) you can do
$$\left( \frac{13}{10} \right)^{-x} =\left(e^{\ln \frac {13}{10}}\right)^{-x} \\
=\left(e^{-x\ln \frac {13}{10}}\right)\\
=e^{x\ln(10/13)}$$
A: You are correct that:
$e^{\ln (\frac{10}{13}x)}=\frac {10}{13}x $.
But
$e^{(\ln \frac {10}{13})x}=(e^{\ln \frac {10}{13}})^x=(\frac {10}{13})^x=(\frac {13}{10})^{-x} $.
You aren't missing any concept.  You are just not reading the question as the person who wrote it intended.  Which, arguably, could be the authors fault.
