How to solve the equation $e^{3x-1} = 5e^{2x}$? Solve the Equation -
$$e^{3x-1} = 5e^{2x}$$
I want to solve this by $\ln$ both sides 
$\ln e^{3x-1} = \ln 5e^{2x}$ 
$ 3x-1 \ln e = 2x \ln5e $ 
$\frac{3x-1}{2x} = \ln 5e$
I then go on to solve for $x$ but didn't get the answer . What went wrong ? 
 A: Here, I'm using the property: $a^b=a^c \Leftrightarrow b=c $ where $a > 0$ and $a\ne1$. Here's how it goes:
$$
e^{3x-1} = 5e^{2x}\implies\\
e^{3x-1} = e^{\ln{5}}\cdot e^{2x}\implies\\
e^{3x-1} = e^{\ln{5}+2x}\implies\\
3x-1 = \ln{5} + 2x\implies\\
3x-2x= \ln{5}+1\implies\\
x= \ln{5}+1\implies\\
x= \ln{5}+\ln{e}\implies\\
x= \ln{(5e)}
$$
Your mistake was that $2x$ is the exponent only on $e$, while according to the rule $\log_{a}{b^c}=c\log_{a}{b}$, the exponent should be on the entire expression of the logarithm before it can be pulled out front. You thought that $\ln 5e^{2x}$ was $\ln (5e)^{2x}$, but that's not a correct interpretation of mathematical symbology. So, you can't really bring that $2x$ out front like you did in your solution. And that's where problems arose.
A: Write $5e^{2x}$ as $e^{ \ln(5)}e^{2x}=e^{\ln(5)+2x}$
A: 
$\ln ab = \ln a + \ln b$, $\ln b^a = a \ln b$.

You seem to have confused $\ln 5 \times e^{2x}$ with $\ln (5e)^{2x}$. The first would simplify according to the first rule, as $\ln 5 + 2x$, while the second would simplify as $2x \ln (5e)$. You should have obtained the first expression, and not the second, which lead to the incorrect answer.
After taking $\ln$ then, you should get $3x - 1 = 2x + \ln 5$, which would give you $x = \ln 5 + 1$, which should be correct.
