# An Inequality about power numbers

Which number is larger? $4^{25}$ or $9^{15}$. Why? I know that it used powers of 2 and 3 but how?

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We can do this in our heads. Simplify to a power of 2 vs. a power of 3, then take out (taking a root) the greatest common divisor of the exponents. When you have done that, you'll have only a couple of two digit numbers to compare. – hardmath Jul 17 '14 at 19:43

$4^{25} = 2^{50} = 32^{10} > 27^{10} = 3^{30} = 9^{15}$ So first one larger.

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Very good. Thanks. – bigli Jul 17 '14 at 19:45

We want to compare $4^{25}$ and $9^{15}$. Rewrite as $2^{50}$ and $3^{30}$. Now, we have that $$\frac{2^{50}}{3^{30}}>1\iff \frac{2^5}{3^3}>1 \iff \frac{32}{27}>1$$ and since the latter is true, we have $2^{50}>3^{30}$.
Note that the first equivalency holds because $\sqrt[10]{\cdot}$ is a monotone function.

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It's not hard to see that $$4^5=1024>9^3=729$$ and now raise to the power $5$.

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Actually $4^5 = 1024$ and $9^3 = 729$ (though of course this doesn't affect the conclusion). – David Z Jul 18 '14 at 3:10
LoL it's an incomprehensible error I do not know how I did it. Thanks. – user63181 Jul 18 '14 at 5:40

$log(4^{25})=50log2=50(.3)=15 \\$(approximately) $$\\$$ $log(9^{15})=30log3=14.31$(approximately)

So, 1st one is bigger.

$$\\$$

Alternatively, $(4^{25})=(4^5)^5$ and $(9^{15})=(9^3)^5$ $$\\$$ $4^5>9^3$

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Please without using $\log$. – bigli Jul 17 '14 at 19:42
You've made a mistake in the 2nd approximation anyway. – hardmath Jul 17 '14 at 19:44
Also, I think you meant $\log{9^{15}} = 30 \log{3}$ not $45 \log{3}$ – suncup224 Jul 17 '14 at 19:44