As the title suggests, my log skills are pretty lacking. Need to learn how to get from $4\log_2(n)=n$ to $n=16$ ($\log$ base $2$). I've searched Google and it seems I am missing some core concept here. I really appreciate the help or guidance.
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Hints: $a \log (n) = \log(n^a)$; $\log_b (n) = x \rightarrow n = b^x$. I assume you are solving in $\mathbb{N}$, that is, the only answers you care about are ones that are positive integers. – MT_ Feb 01 '14 at 20:43
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You need to use the Lambert W-function, which can't be solved with elementary functions (but can be expressed as an infinite sum using the Lagrange inversion forumula)
However just by trying random powers of 2 you can get an integer solution of $n = 16$, probably the easiest way.
qwr
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Hint: $4\log_2(n) = n$ is the same as $n^4 = 2^n$. As $n=16$ is indeed a solution, the only question left is whether there is any other natural number that satisfies that equality. But, $2^n$ (eventually) grows faster than $n^4$ as $n$ grows (more precisely, $2^n > n^4$ for all $n$ bigger than $16$ - induction or calculus can show this).
ir7
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$4\log_2n=n\iff2^{4\log_2n}=2^n\iff n^4=2^n\iff n=2^k$ , since $n\in\mathbb{N}\iff2^{4k}=2^{2^k}\to$ $\to4k=2^k\iff k=2^{k-2}$. Can you take it from here ? :-)
$4\log_2n=n\iff4\dfrac{\ln n}{\ln 2}=n\iff\dfrac{\ln n}n=\underbrace{\dfrac{\ln2}4}_\text{constant}.\quad$ Now draw the graphic of $\dfrac{\ln x}x$ , and tell me what you notice. :-)
Lucian
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