TITLE: How many digits are in $125^{100}$?
QUESTION [8 upvotes]: What I can think of thus far is that $125^{100} = (\frac{1000}{8})^{100} = \frac{1000^{100}}{2^{300}}$
I know that $2^{10} = 1024$ so $\frac{1000^{100}}{1024^{30}}$.
That's all I can figure out this far. 
I was thinking to divide the numerator and denominator of $\frac{1000^{100}}{1024^{30}}$ by $1000^{30}$ and I think that would give me $\frac{1000^{70}}{1.024^{30}}$ but I'm not even sure if this is correct. 
Can someone please help me solve this?
Edit: How can I solve this without the use of logarithms?

REPLY [2 votes]: As you worked out, $125^{100}=\dfrac{10^{210}}{1.024^{30}}$, so it is enough to show $1<1.024^{30}<10$ to conclude there are $210$ digits. But the left inequality is obvious and as $1+x\leqslant e^x$, we get $1.024^{30}\leqslant e^{0.024\times 30}=e^{0.72}<e<10,\;$ so the right inequality holds true as well.