Solving equations with exponentials and a non-exponential term. I know how to solve exponential equations. Just use logarithms, e.g.,
$$
2^x-3=0 \\
2^x=3 \\
x=log_23 \\
$$
But on a recent math test I found an equation of the form:
$$
2^{n-3}=\frac {20}{n}
$$
Which can be rearranged to,
$$
2^{n-3}n-20=0\\
$$
Now using graphing calculator the solution turns out to be $n=5$, and I don't know if the calculator figured this out algebraically or used some kind of iterative method like the Newton–Raphson method for square roots. But the point is: how on earth do you solve these equations with algebra?
There are other equations similar to this which are products and sums of different expressions e.g., exponentials, quadratics etc. which can be solved on their own but when together are much harder (for me) to solve, e.g.,
$$
2^nn+c=0\\
2^n+n=0\\
2^xx^2=0
$$
I could go on forever, but how are these equations that seem impossible to solve solved? I can remember how I used to have no idea how to solve quadratics because I tried to isolate $x$ but I couldn't because I was going about it the wrong way. Of course now quadratics are easy to for me to solve.
So how are equations like $2^{n-3}n-20=0$ solved using algebra, or can they be? Or in other words how are the roots of functions such as $f(x)=2^{n-3}n-20$ found?
P.S. I did browse for an answer to this question but I didn't really know how to phrase it when searching, most of the searches I tried came up with solutions to different problems I wasn't interested in. I don't really know what to call equations like the ones discussed in this question and therefore it is difficult to make a search on how to solve them.
 A: You can modify your equation as such
$$ \ln(2)n e^{\ln(2)n} = 160\ln(2) $$
So that
$$ n = \frac{1}{\ln(2)} W(160\ln(2))=5$$
Where $W$ is the Lambert W function. I do not know any identity on W that would lead the result 5 without numerically computing W (which is usually done using root-finding).
A: Suppose $n\cdot 2^n=160$.
Since $160=2^5\cdot 5$, you know that $5$ divides $n$, so $n=5m$. Then
$$
5m\cdot2^{5m}=160
$$
becomes $m\cdot 2^{5m}=32$ and so $m=1$.
However, the general solution of $x\cdot 2^x=a$, for arbitrary $a$, cannot be determined “explicitly”, without using “higher level”  functions such as Lambert's $W$.
A: Although the Lambert W function is a perfectly valid solution, as described in the other answers, it is impossible to compute with a standard calculator, and is beyond your current math level anyway (judging from your question).
The equation $2^{n-3}=20/n$ tells you that $20/n$ is a multiple of $2$. Furthermore, it should be obvious that $n>3$, since otherwise $2^{n-3}\le 1$ but $20/n>1$. Even better, since $20=2\times 2\times 5$, you only have to check $n=4$ and $n=5$. But $20/4=5$ which is not a multiple of $2$. Hence the solution must be $n=5$.
A: In a general manner, any equation which can write $$f(x)=A+Bx+C\log(D+Ex)=0$$ has solutions in terms of Lambert function.
If, for any reason, you cannot use it, only numerical methods will find the root. Probably, the simplest should be Newton which, starting from a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For illustration purposes, let us find the zero of $$f(x)=\pi^x+\sqrt{5}x-1234=0$$ A quick plot of the function will show you that the root is close to $6$; so, let us use $x_0=6$; then Newton scheme will provide the following iterates : $$x_1=6.23504003679589$$ $$x_2=6.20859973003907$$ $$x_3=6.20819208155631$$ $$x_4=6.20819198656474$$ $$x_5=6.20819198656473$$ which is the solution for fifteen significant digits.
