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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?

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marked as duplicate by Rory Daulton, YuiTo Cheng, max_zorn, StammeringMathematician, Lord Shark the Unknown May 12 at 6:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Already asked zillion times. Rewrite in the form $\frac{\log x}x<\frac{\log y}y$. $\endgroup$ – Yves Daoust May 11 at 20:18
  • $\begingroup$ Think about the function $\frac{ln(x)}{x}$ $\endgroup$ – Gabi G May 11 at 20:18
  • $\begingroup$ but how do i know which one is less than the other? @YvesDaoust $\endgroup$ – user130306 May 11 at 20:20
  • $\begingroup$ Hint: increasing or decreasing function ? $\endgroup$ – Yves Daoust May 11 at 20:21
  • $\begingroup$ 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). $\endgroup$ – Rory Daulton May 12 at 0:10
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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:

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

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  • $\begingroup$ from where are you getting the blue part that you added on? $\endgroup$ – user130306 May 11 at 22:56
  • $\begingroup$ I didn't add it - the first one is from writing $2019$ as $(2018 + 1)$ and using the distributive law. The second blue part is obvious. $\endgroup$ – MarianD May 11 at 23:39
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Hint: $f(x)=\ln x/x$ is decreasing for $x>e$.

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  • $\begingroup$ im sorry im just still confused. are you saying i should do $\log 2018/ 2019 < \log 2019/2018$? $\endgroup$ – user130306 May 11 at 22:56
  • $\begingroup$ sorry, I finally understood what you meant. thank you $\endgroup$ – user130306 May 12 at 18:19
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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.

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