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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.

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    $\begingroup$ $n2^n=160$. You can find a solution by inspection. Generally, such equations can't be solved explicitly. $\endgroup$ – David Mitra May 17 '15 at 9:55
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    $\begingroup$ en.m.wikipedia.org/wiki/Lambert_W_function may help with some $\endgroup$ – bnosnehpets May 17 '15 at 9:59
  • $\begingroup$ @DavidMitra What is inspection? Is that just guess and check? $\endgroup$ – hddh May 17 '15 at 10:01
  • $\begingroup$ Yes.${}{}{}{}{}$ $\endgroup$ – David Mitra May 17 '15 at 10:02
  • $\begingroup$ @DavidMitra How did the calculator do it then? Using advanced guess and check? $\endgroup$ – hddh May 17 '15 at 10:05
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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 $W$.

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  • $\begingroup$ Thanks, this solution was simple. I'm guessing the answer was a convenient whole number because it was from a test question and they tend to have convenient numbers throughout. I'm interested by this Lambert W function. Could you reference some useful teaching resources to get me started (I checked the wikipedia article and I don't understand much). $\endgroup$ – hddh May 17 '15 at 10:33
  • $\begingroup$ @hddh You can easily show that the solution of $x\cdot 2^x=a$, with $x>0$ is unique whenever $a>0$, by rewriting it as “find the zeros of the function $f(x)=\log x+x\log 2-\log a$”. You can prove that the function is strictly increasing and takes on negative and positive values, so it meets the $x$-axis exactly once. Lambert's $W$ function is the inverse of $x^x$, defined on $(1/e,\infty)$. $\endgroup$ – egreg May 17 '15 at 10:37
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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).

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  • $\begingroup$ Could you explain how you got from $2^{n-3}n-20=0$ to $\ln(2)n e^{\ln(2)n} = 160\ln(2)$. This seems above my level anyway. $\endgroup$ – hddh May 17 '15 at 10:14
  • $\begingroup$ Firstly $2^{n-3}=2^n/8$, $8\times 20=160$ and $2^n=e^{\ln(2)n}$. The $\ln(2)$ has been added on both sides of the equation in order to use the Lambert W function, which is the solution to $y=x e^x$ given $y$ $\endgroup$ – vanna May 17 '15 at 10:23
  • $\begingroup$ Ahh, cheers. I'll have to research this Lambert W function. Any resources to get me started? (OTHER than wikipedia lol). $\endgroup$ – hddh May 17 '15 at 10:28
  • $\begingroup$ For special functions online references like Wikipedia or Wolfram are not as bad imho $\endgroup$ – vanna May 17 '15 at 10:33
  • $\begingroup$ @vanna: $n = \frac{1}{\ln(2)} \operatorname{W}(160\ln(2))=5$ is exact, since $\operatorname{W}(160\ln(2))=\operatorname{W}(5\ln(2)\exp(5\ln(2)))=5\ln(2)$. $\endgroup$ – g.kov Jun 13 '15 at 7:00
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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 $n=5$.

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  • $\begingroup$ Yeah, this is one of the simpler ways, it seems, to go about it; by reasoning through it, but in terms of isolating $n$ in the equation just like you would isolate $x$ in a quadratic, that is what I'm interested in. Thanks for your answer, If I put questions up on this website I get answers within minutes sometimes! $\endgroup$ – hddh May 28 '15 at 11:11
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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.

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  • $\begingroup$ Thanks, how do you think the calculator did it? Did it use an approximation method or otherwise? $\endgroup$ – hddh May 28 '15 at 11:13
  • $\begingroup$ No idea ! What kind of calculator do you use ? $\endgroup$ – Claude Leibovici May 28 '15 at 11:48

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