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We want to find the least $n$ that satisfies this inequality: $$ \frac{4 n}{ \log (4 n)} > \frac{90 n \log (113)}{113 \log (3 n)}. $$
We divide each side by $\frac{4 n}{ \log (3 n)}$ to move $n$ to LHS and $4$ to RHS, $$ \frac{ \log (3 n)}{ \log (4 n)}>\frac{45 \log (113 )}{226}. $$ Wolfram Alpha and Mathematica reduce to, $$ n>3^{\frac{226}{45 \log (113)-226}} e^{-\frac{90 \log (2 ) \log (113)}{45 \log (113)-226}}, $$ but I have no idea how to get there.

Are there any logarithmic identities that would help?

Fred Daniel Kline
  • 1,348

2 Answers2

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You should be able to separate a $\log n$ from everything else by multiplying both sides by $\log 4n$ and then using $\log ab = \log a + \log b$. From there, just solve for $\log n$ and then just $n$.

AlgorithmsX
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Assuming $n\ge 1$ and calling $A=\frac{226}{45 \log (113 )}\approx 1.062$ we have

$$ \begin{align} \frac{4 n}{ \log (4 n)} > \frac{90 n \log (113)}{113 \log (3 n)} & \iff A \log 3n > \log 4n \\ & \iff \log (3n)^A > \log 4n \\ & \iff (3n) ^A>4n\\ & \iff n^{A-1}> 4/3^A \\ & \iff n>(4/3^A)^{1/(A-1)}\approx 33.5855 \end{align} $$

Hence the least integer that verifies the inequation is $n=34$

leonbloy
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