What is the most systematic way to do this problem? I used the definition of logarithms and brute force to find $n = 16$, but I feel as though that was the worst way possible.
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what $\log$ it is? – Dr. Sonnhard Graubner Oct 03 '17 at 09:55
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The base is 2 . – darylnak Oct 03 '17 at 09:56
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Equations that have both $\log n$ and $n$ in them are usually not solvable without brute force. This one you can simplify to $4\log_2n=n$, but after that, you're out of luck. – Arthur Oct 03 '17 at 09:58
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Assuming $n\neq 0$, you may simplify by dividing both sides by $8n$ and then take base 2 on both sides to get $n = 2^{\frac n4}$. To my knowledge there is not much more you can do here. A graphical solutuion or a guess + showing that its the only solution would do it. – Wisław Oct 03 '17 at 10:00
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Are you looking for integer solutions or any positive real solution? – Robert Z Oct 03 '17 at 10:12
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After dividing by $n$ (and thereby removing the solution $n=0$) and simplifying, we get $$ -\log(n)e^{-\log(n)}=-\frac{\log(2)}4 $$ Therefore, $$ \begin{align} n &=e^{-\operatorname{W}\left(-\frac{\log(2)}4\right)}\\[6pt] &=-\frac4{\log(2)}\operatorname{W}\left(-\frac{\log(2)}4\right) \end{align} $$ For negative arguments, Lambert W has two real branches giving two solutions: $1.2396277295227621418$ and $16$.
robjohn
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write your equation in the form $$n\left(4\frac{\ln(n)}{\ln(2)}-n\right)=0$$ you can choose a numerical way or the LambertW function. $$n_1=-4\,{\frac {{\rm W} \left(-1/4\,\ln \left( 2 \right) \right)}{\ln \left( 2 \right) }} $$ $$n_2=-4\,{\frac {{\rm W} \left(-1,-1/4\,\ln \left( 2 \right) \right)}{\ln \left( 2 \right) }} $$ $$n=0$$ is impossible
Dr. Sonnhard Graubner
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Don't you think that $-4,{\frac {{\rm W} \left(-1,-1/4,\ln \left( 2 \right) \right)}{\ln \left( 2 \right) }}$ is a really complicated way of writing $16$? – orlp Oct 03 '17 at 10:05
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This is a good answer (and the one I was writing up, but he beat me to it). The point is that $W$ allows you to solve the equation fully, without needing to resort to other heuristics. It's easy enough to evaluate $W$ and get 16. – Mr. Chip Oct 03 '17 at 10:09
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ok you are welcome, i don't want to beat you, we all should help to solve mathematical Problems from others – Dr. Sonnhard Graubner Oct 03 '17 at 10:13
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First of all, I would think one could also call $n=0$ a solution, because of $\lim_{n\to0}n\log_2(n)=0$. But discarding this solution and recalling that the question was whether we can do without brute force, let me suggest to divide the equation by $8n$ which gives $\log_2(n)=n/4$. Antilog leads to $n=2^{n/4}$. Raising to the power 16 finally leads to $n^{16}=16^n$. This makes it quite obvious that $n=16$ is a solution, albeit not the only one.
Gerhard S.
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