Compare two powers

Question

how can I compare these powers: $3^{500}$ and $5^{300}$ What I did is:

$\log_3(3^{500})$ and $\log_3(5^{300})$ So I have

$500$ and $\log_3(5^{300})$ Now I do not know what to do. Thank you in advance!

I would suggest using a base that is available directly on your calculator, base $10$ or base $e$ (natural logarithm). Whatever base you use, $\log(3^{500})=500\log 3$. But because these numbers are very special, there is a shortcut. — André Nicolas, Mar 03 '14 at 16:33

score 2 · Accepted Answer · answered Mar 03 '14 at 16:35

2

No need of messing things up with logarithms: since clearly $\;3^5>5^3\;$ and$$500=5\cdot 100\;,\;\;300=3\cdot 100\implies$$

$$3^{500}=\left(3^5\right)^{100}>\left(5^3\right)^{100}=5^{300}$$

answered Mar 03 '14 at 16:35

DonAntonio

Is this a property that if the exponential 5 is bigger than the base 3 then the power 5^3<3^5 ? – wonderingdev Mar 03 '14 at 16:39
@John, just calculate by hand! $;3^5=243;,;;5^3=125;$ ... – DonAntonio Mar 03 '14 at 16:40
:) Thank you , excuse me it was a stupid question – wonderingdev Mar 03 '14 at 16:42

score 0 · Answer 2 · answered Mar 03 '14 at 16:35

0

Take the one hundreth root of both numbers. Then you're left with comparing $3^5=243$ and $5^3=125$.

answered Mar 03 '14 at 16:35

score 0 · Answer 3 · answered Mar 03 '14 at 16:39

0

$\sqrt3\simeq1.7\iff3\sqrt3\simeq5\iff\log_35^{300}=300\cdot\log_35\simeq300\cdot\log_3\Big(3\sqrt3\Big)=300\cdot\Big(1+\frac12\Big)$ $=300+150=450$.

answered Mar 03 '14 at 16:39

Lucian

3 Answers3