Simplify $\ln(e^{2x+1})$

Question

I was trying to do an integration problem with u sub and got stuck, one part of the equation was this $\ln(e^{2x+1})$

this is suppose to simplify really nicely according to a site, is there a rule for this i get that $\ln(e) =1$ and that $\ln(e^{x})=x$. So would $\ln(e^{2x+1})=2x+1$? Is this a rule that I may have forgotten about? I'm more interested in the properties behind this if it is a rule, just out of curiosity as to why it is so.

Answer 1

Think it this way: $\ln(e^{f(x)})$ defines the exponent such that if you take $e$ to the power of that exponent, will equal to $f(x)$. So, what power does $e$ need to be raised to for it to equal to $e$ to the power of $f(x)$? $f(x)$ itself, of course, and this works for any exponent since we made no assumptions on it.

Answer 2

Yes, because the natural logarithm(ln) of a number is its logarithm(log) to the base of the mathematical constant e.

Answer 3

$$\ln(e^{2x+1})=2x+1$$ Take exponential on both side $$e^{\ln(e^{2x+1})}=e^{2x+1}$$ but $e^{\ln a}=a$ , so: $$e^{2x+1}=e^{2x+1}$$ You have these rules $$ \begin{cases} e^{\ln (f(x))}=f(x) \\ \ln(e^{f(x)})=f(x) \end{cases} $$