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I am trying to show algebraically that $8^3>9^{8/3}$. This came from trying to complete the base case of an induction proof.

I have struggled because $8$ and $9$ cannot be manipulated to be the same base. Otherwise I could just argue that $3>\dfrac{8}{3}$.

I tried raising both sides to the third power and got $8^9>9^8$. I can rewrite this as $8^9>9^{9-1}$ but I am not sure if this is the right direction.

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  • $\begingroup$ Related: math.stackexchange.com/questions/410697, math.stackexchange.com/questions/1406965 $\endgroup$
    – player3236
    Commented Jan 15, 2021 at 5:21
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    $\begingroup$ One way [probably more exotic than you want] is to rewrite it as $$8 > \left(1 + \frac{1}{8}\right)^8$$ which is true since $$\left(1 + \frac{1}{n}\right)^n \uparrow e < 8$$ $\endgroup$
    – Brian Moehring
    Commented Jan 15, 2021 at 5:22
  • $\begingroup$ Why can't you just calculate $8^9$ and $9^8$? They're big numbers, but they're not even bigger than one billion. $\endgroup$
    – Duncan Ramage
    Commented Jan 15, 2021 at 5:33
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    $\begingroup$ @DuncanRamage This question came up during a tutoring session. The student I was tutoring was concerned that it would take too long to compute by hand if a similar question were to appear on an exam. $\endgroup$
    – FoiledIt24
    Commented Jan 15, 2021 at 5:34
  • $\begingroup$ An approach specific to this problem is that we have $8^9>9^8$ iff $(\frac{8}{9})^8>\frac{1}{8}$, which is true as $(\frac{8}{9})^8=(\frac{8}{8+1})^8=\frac{8^8}{8^8+8\cdot 8^7+...+1}>\frac{1}{8}$ because $8^8> {8\choose k}8^{8-k}1^k$ for $k>1$. $\endgroup$
    – Yuqiao Huang
    Commented Jan 18, 2021 at 5:32

2 Answers 2

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We want to compare $x^y $ vs. $y^x$, for $x, y > e$. Take log base $y$ of both sides, we get $y \log_y(x) = y \ln(x) / \ln(y)$ vs. $x$.

Note that if $x = y$, then $y ln(x) / ln(y) = x$. Consider $y$ as a constant and take derivatives. $\frac{y}{x ln(y)}$ vs $1$.

Therefore, if $x > y$, then $\frac{y}{x \ln(y)} < 1$ (Here is where we use $y > e$, so that $ln(y) > 1$. Without this assumption, it could be possible that $\frac{y}{x \ln(y)} > 1$). Therefore, 1 is bigger. Thus, for $x > y$, $x > y \ln(x) / \ln(y)$, so $y^x$ is bigger than $x^y$.

In general, $small^{big} > big^{small}$ for numbers $> e$

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An approach that is alternative to (but inferior to) David Lui's answer is if you happen to know that

$$\log_{10} ~2 \approx 0.301 ~~\text{and}~~ \log_{10} ~3 \approx 0.477.$$

Then, you simply compare

$$0.301 \times (3 \times 3) ~~\text{vs}~~ 0.477 \times [2 \times (8/3)].$$

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