1
$\begingroup$

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?

3 Answers 3

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:

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

2

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

1

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