My thoughts:






  • 2
    $\begingroup$ It should be $-\frac{n}{50} + \log\left(1 + \frac{n}{50}\right) = \log(0.05)$. The log of a sum is not the sum of logs. $\endgroup$ – Winther Jul 14 '15 at 18:07
  • $\begingroup$ You might want to use en.wikipedia.org/wiki/Lambert_W_function ... $\endgroup$ – Stephan Kulla Jul 14 '15 at 18:07
  • $\begingroup$ Yeah, I think the Lambert omega function's gonna be needed, as I ran into the ground trying to do it without that $\endgroup$ – Alan Jul 14 '15 at 18:27
  • $\begingroup$ This question at the level you are asking it (precalculus) do not have an analytical solution: there is no simple formula for $n$. As the answers below say one can 'make up' a function that solves it, but that function is highly non-trivial and far beyond the precalculus level (and imo not very useful). That is not to say that the equation has no solutions, it does. One can prove that it has two real solutions and there are many methods to find them numerically, see forexample this question. $\endgroup$ – Winther Jul 15 '15 at 2:14

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:


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.

  • 1
    $\begingroup$ There are two real solutions to this problem. The second solution is 'hidden' in the fact that the W-function is multi-valued. $\endgroup$ – Winther Jul 15 '15 at 2:03
  • $\begingroup$ Yes, in peace with the fact that we don't have to face complex roots I forgot to mention that for negative values we have two real branches of this function @Winther $\endgroup$ – Renato Faraone Jul 15 '15 at 12:12

$$e^{-\frac{n}{50}}+\frac{n}{50}e^{-\frac{n}{50}}=0.05 = e^{-\frac{n}{50}}\left(1 + \frac{n}{50}\right) = 0.05$$ 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?


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 $2.5=e^{-\frac n {50}}(n+50)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.