Solving logarithm equation $$e^{-\frac{n}{50}}+\frac{n}{50}e^{-\frac{n}{50}}=0.05$$
My thoughts:
$$\ln(e^{-\frac{n}{50}}+\frac{n}{50}e^{-\frac{n}{50}})=ln(0.05)$$
$$\ln(e^{-\frac{n}{50}})+\ln({\frac{n}{50}e^{-\frac{n}{50}}\over{e^{-\frac{n}{50}}}})=\ln(0.05)$$
$$\ln(e^{-\frac{n}{50}})+\ln(\frac{n}{50})=\ln(0.05)$$
$$-\frac{n}{50}+\ln(\frac{n}{50})=\ln(0.05)$$
$$-\frac{n}{50}+\ln(n)=\ln(2.5)$$
 A: $$e^{-\frac{n}{50}}+\frac{n}{50}e^{-\frac{n}{50}}=0.05 = e^{-\frac{n}{50}}\left(1 + \frac{n}{50}\right) = 0.05$$ Take the natural logarithm of both sides to get $$\ln \left(1 + \frac{n}{50}\right) -\frac{n}{50} = \ln 0.05$$
Can you take it from there?
A: This is not a pre-calculus answer but I'll try to explain how those kind of problem are usually solved:
first of all we set $x=\frac n{50}$ so we can rewrite the equation in a more clear form:
$$xe^{-x}+e^{-x}=\frac 1{20}$$
$$(x+1)e^{-x}=\frac 1{20}$$
Now we multiply all by $-\frac 1e$ so we can get to:
$$(-1-x)e^{-1-x}=-\frac 1{20e}$$
To solve this we must introduce a new function called Lambert W (Omega) function defined as follow:
$$W(x)e^{W(x)}=x$$
And this lead us to:
$$-1-x=W(-\frac 1{20e})$$
Substitution gives:
$$n=-50-50W(-\frac 1{20e})$$
Note that argument of the function is greater than $-\frac 1e$ so we are not in trouble with complex values.
A: This is not a complete answer,  but I felt it was worth saving to show some thought processes, so here's some food for thought:
Generally you want a product, not a sum, when you are wanting to take a natural log.  So,  factoring your left hand side, you get $e^{\frac {-n} {50}}(1 + \frac n {50})=.05$.  Taking natural logs, we get 
$\ln (e^{\frac {-n} {50}}(1 + \frac n {50}))=\ln (.05)=\ln (e^{\frac {-n} {50}})+\ln (1+ \frac n {50})=-\frac n {50}+\ln (1 + \frac n {50})$
Now, again we don't want a sum in the natural log, so convert that to a fraction and we get 
$\ln (.05)=-\frac n {50}+\ln (\frac {n+50} {50})=-\frac n {50}+\ln (n+50)-\ln (50)$
Adding the $\ln (50)$ to the other side gives us 
$-\frac n {50}+\ln (n+50)=\ln (.05)+\ln (50)=\ln (.05 * 50)=\ln (2.5)$
Now, at this point we should pause and make sure we don't have any sign confusion/errors, since you can't take natural log of negative numbers. From the original equation factored, we see that since $.05>0$ and $e^{\frac {-n} {50}}>0$, we must have $1+\frac n {50}>0$,  so $n>-50$. Thus we have no problems with our above expression being undefined.
Now, we want to combine things in our last equation, so we want $-\frac n {50}$ as a natural log,  so that's $-\frac n {50}=\ln (e^{-\frac n {50}})$  Plugging that in to the last equation, we have 
$\ln (e^{-\frac n {50}})+\ln (n+50)=\ln (2.5)=\ln (e^{-{\frac n {50}}}(n+50))$
So $2.5=e^{-\frac n {50}}(n+50)$
