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Which is bigger between $2018^{2019}$ or $\ 2019^{2018}\ $?

When taking logs of both sides and I get:

$2019\log(2018)\ $ and $\ 2018 \log(2019)$

I know $\log 2019\gt \log 2018$ so does this mean that $2019^{2018}$ is the biggest one? And did I do it properly?

user130306
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    Already asked zillion times. Rewrite in the form $\frac{\log x}x<\frac{\log y}y$. –  May 11 '19 at 20:18
  • Think about the function $\frac{ln(x)}{x}$ – Gabi G May 11 '19 at 20:18
  • but how do i know which one is less than the other? @YvesDaoust – user130306 May 11 '19 at 20:20
  • Hint: increasing or decreasing function ? –  May 11 '19 at 20:21
  • Rewrite your question this way: Which is larger, $2018^{1/2018}$ or $2019^{1/2019}$, and this question has been asked here before (with different numbers but the same method of solution). – Rory Daulton May 12 '19 at 00:10
  • @RoryDaulton sorry, I'm just failing to see how $2018^{2019}$ is the same as $2018^{\frac{1}{2018}}$ – user130306 May 12 '19 at 00:37
  • Take the $2018 \cdot 2019$th root of each side of your equation and you end up with my equation. (I.e. take each side to the power of $\frac 1{2018\cdot 2019}$.) – Rory Daulton May 12 '19 at 00:40

3 Answers3

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Hint: $f(x)=\ln x/x$ is decreasing for $x>e$.

operatorerror
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No, $\log 2019>\log 2018$ is simply equivalent to $2019>2018$ by monotonicity of the logarithm, and this does not prove the claim.

1

Your idea to take logarithm of both expressions is good. Now \begin{align} 2019\cdot\log 2018&=\color{blue}{2018\cdot\log 2018}+\log2018\tag1\\[2em] 2018\cdot\log 2019 &=2018\cdot\log \left(2018\cdot{2019\over2018}\right) \\ &= 2018\cdot(\log 2018 + \log{2019\over 2018}) \\ &=\color{blue}{2018\cdot\log 2018}+2018\cdot\log\left({2019\over 2018}\right)\tag2 \end{align}

As both $(1)$ and $(2)$ have their first addend (in blue) the same, what is greater:

$$\log2018,\ \text{or}\tag3$$ $$2018\cdot\log\left({2019\over 2018}\right)\ ?\tag4$$

MarianD
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