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.
Best Answer
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 $W$.