How do I solve the equation $e^{\ln(2x+1)} = 5x$? The problem is 
$$e^{\ln(2x+1)}    =5x$$
I've tried using natural logs to both sides like..
$2x+1= \ln 5x $
But I'm not sure if $\ln$ and $e^{\ln}$ cancel out.
 A: Yes, the exponential and the logarithm are inverses: $$e^{\ln(\color{red}{\rm stuff})} = \color{red}{\rm stuff}, \quad \ln(e^{\color{blue}{\rm stuff}}) = \color{blue}{\rm stuff},$$ so that $e^{\ln(2x+1)} = 2x+1$.
A: e raised to the power of a natural log cancels out both. This just leaves you with $2x+1=5x,$ or $x=\frac{1}{3}$
A: The logarithm is the inverse of the exponential, meaning $exp \circ log = id$.
It then follows that $e^{log(2x+1)}=2x+1=5x$ which should be easy to solve.
A: $e^{ln(x)}$ $=$x
check the following pdf for this and more properties of logs
http://assets.wne.edu/95/exps_and_logs.pdf
Using this property we can see that
$2x+1$ = $5x$
$x$ $=$ ${1/3}$
A: If you apply natural log to both sides, you should instead get$$\begin{align*}
\ln\left[e^{\ln(2x+1)}\right] &= \ln(5x)\\
\ln(2x+1)\cdot\ln(e) &= \ln(5x)\\
\ln(2x+1)\cdot1 &= \ln(5x)\\
\ln(2x+1) &= \ln(5x)\\
&\vdots
\end{align*}$$
A: $e^{\ln a}=a$, for $a\gt-1/2$. so $2x+1=5x$ and solve for $x$. 
A: Because $e^x$ and $\ln x$ both have domains $x > 0$ where they are injective then, given that they are inverses of one another means that the composition of them yields the identity function. Specifically we have $$\ln e^x = e^{\ln x} =x$$
The same logic applies to other functions, for example, consider the function $x^3$ and the function $x^{1/3}$, they are inverses of another. The cube function is the inverse of the cube root function and vice-versa. So composing them yields the identity again $$(x^3)^{1/3} = (x^{1/3})^3 = x $$
Another example would be, considering a suitable domain for $x$, that $$\arcsin \sin x = \sin \arcsin x = x$$
The list of example goes on... and I hope I've provided a valuable insight into your understanding of how composing inverses yields the identity. 
