# How to know which value is bigger? [duplicate]

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?

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

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

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:

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

• from where are you getting the blue part that you added on? – user130306 May 11 at 22:56
• 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. – MarianD May 11 at 23:39

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

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

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