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So I have a question which is:

Which is larger?

$$2.2^{3.3} \text{ or } 3.3^{2.2} $$

Now I need to find out with using a calculator but the answer is $3.3^{2.2}$.

The only thing I could think of is rounding.

So you know:

$2^3=8$ and $3^2=9$

I'm interested in seeing if there are other ways just because there might be the possiblity of being asked:

Which is larger?

$$2.5^{3.5} \text{ or } 3.5^{2.5} $$

So if I use the rounding idea, would I just round normally. I would get:

$$ 3^4 \text{ or } 4^3 $$

which shows $3^4$ is larger.

This is the only way I could think of, are there any other ways without using a calculator to determine which is larger?

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    $\begingroup$ The square root function is increasing, so you can ignore the $0.5$ and it's enough to compare $\left(\dfrac{25}{10}\right)^3$ and $\left(\dfrac{35}{10}\right)^2$, or equivalently $25^3$ and $10\cdot 35^2$, which you can easily do by hand. $\endgroup$
    – Git Gud
    Commented Sep 2, 2015 at 18:27
  • $\begingroup$ Noting once again in passing that I despise "without a calculator" questions. There is no magic pixie dust in the calculator that couldn't be replicated, in principle, by hand. $\endgroup$
    – user7530
    Commented Sep 4, 2015 at 6:47

4 Answers 4

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$$2.2^{3.3}\gtrless3.3^{2.2}$$ $$(2.2^3)^{1.1}\gtrless (3.3^2)^{1.1}$$ $$2.2^3\gtrless 3.3^2$$ $$(2\cdot 1.1)^3\gtrless (3\cdot 1.1)^2$$ $$2^3\cdot 1.1\gtrless 3^2$$ $$8\cdot 1.1\gtrless 9$$ $$8.8<9.$$

So, $$2.2^{3.3}<3.3^{2.2}.$$

Can you find now which number is larger: $2.25^{3.375}$ or $3.375^{2.25}$ ;) ?

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    $\begingroup$ That was a beautiful proof. And no 'e' discussion. $\endgroup$
    – JTP - Apologise to Monica
    Commented Sep 2, 2015 at 22:13
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    $\begingroup$ Umm, how did you go from step 4 to 5? It should be $2^3 1.1^3 \gtrless 3^2 1.1^2 \rightarrow 8 \cdot 1.331 \gtrless 9 \cdot 1.21 \rightarrow 10.648 < 10.89$. $\endgroup$
    – geometrian
    Commented Sep 3, 2015 at 15:18
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    $\begingroup$ That step seems to include division by $1.1^{2}$ $\endgroup$
    – njuffa
    Commented Sep 3, 2015 at 15:33
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    $\begingroup$ @Oleg567 Welcome to the "Hot Network Questions" list effect. $\endgroup$
    – Barry
    Commented Sep 3, 2015 at 21:31
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    $\begingroup$ @Oleg567 more like, how did you find this question? :D $\endgroup$
    – Caddy Heron
    Commented Sep 4, 2015 at 2:30
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One nice approach is the calculus approach. In particular: $$ a^b > b^a \iff a^{1/a} > b^{1/b} $$ In order to compare $a^{1/a}$ to $b^{1/b}$, we could consider the function $f(x) = x^{1/x}$ (or its logarithm, $\frac{\ln(x)}{x}$). On which interval is the function increasing? On which interval is it decreasing?

Note that this will only work, however, if both values are either greater than $e \approx 2.718$, or if both are less than $e$ (so, if $a = 3$ and $b=4$). Otherwise, it's not immediately clear.

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    $\begingroup$ Your last paragraph basically says, "Unfortunately, this doesn't actually help you at all." $\endgroup$
    – Akiva Weinberger
    Commented Sep 2, 2015 at 19:19
  • $\begingroup$ @columbus8myhw True but we get quite a lot of this style of question for different values of $a$ and $b$, so I think there's still plenty of value in giving techniques that don't work for this particular $a,b$. $\endgroup$
    – David Richerby
    Commented Sep 3, 2015 at 12:51
  • $\begingroup$ If they are the same side of $e$ then you can use the answers at math.stackexchange.com/questions/517555/… but not here $\endgroup$
    – Henry
    Commented Sep 4, 2015 at 6:06
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Another method is to use the fact that $\log a^b=b\log a$, so that you can reduce the exponents even if they don't have any obvious common divisors:

$$2.2^{3.3}\gtrless3.3^{2.2}$$ $$3.3 \log 2.2 \gtrless 2.2 \log 3.3$$ $$1.5 \log 2.2 \gtrless \log 3.3$$ $$2.2^{1.5} \gtrless 3.3$$ $$2.2 \cdot 2.2^{0.5} \gtrless 3.3$$ $$2.2^{0.5} \gtrless 1.5$$ $$(2.2^{0.5})^2 \gtrless (1.5)^2$$ $$2.2 < 2.25$$ $$\Rightarrow 2.2^{3.3}\lt3.3^{2.2}$$

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Let's calculate the quotient: $$ \frac {3.3^{2.2}}{2.2^{3.3}} = \frac {3.3^{2.2}}{2.2^{2.2} \cdot 2.2^{\fbox{1.1}}} = \frac {(3/2)^{2.2}}{2.2^{1.1}} > \frac {1.5^2}{2.2^{1.1}} = \dots ? $$ EDIT: I edited out the shameful mistake, but now this calculation isn't useful. Oleg fixed it nicely, see below.

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    $\begingroup$ $2.2^{3.3} \neq 2.2^{2.2}\times 2.2$ $\endgroup$
    – yoann
    Commented Sep 2, 2015 at 23:57
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    $\begingroup$ Your idea is right at all, but needs some corrections: $$\dfrac{3.3^{2.2}}{2.2^{3.3}}=\left(\dfrac{3.3^2}{2.2^3}\right)^{1.1}=\left( \frac {(3/2)^2}{2.2} \right)^{1.1}=\left(\dfrac{2.25}{2.2}\right)^{1.1}$$ $\endgroup$
    – Oleg567
    Commented Sep 3, 2015 at 4:12
  • $\begingroup$ @Oleg567 Very sorry, of course, I wrote it after midnight... Thanks! $\endgroup$
    – eudes
    Commented Sep 3, 2015 at 7:56

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