How many digits are in $125^{100}$? 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?
 A: $2^{10}\approx 10^3$, so approximately,  $\frac {1000^{100}}{10^{90}} =\frac {100^{100}\cdot 10^{100}}{10^{90}}=100^{100}\cdot 10^{10}=10^{210}$...  So about $211$.
A: Calculate 
$\log_{10}125^{100}$
$= 100\cdot \log_{10} (1000/8)$
$= 100\cdot(3-3\log_{10}2)$
$= 100\cdot(3-0.9030)$
$= 100\cdot(2.0970)$
$= 209.70$
Therefore number of digits $= [209.70]+1 = 210.$
A: Any solution implicitly computes a logarithm ($\log_{10} 5$ is approximately the answer to this question divided by $300$), but one can correct Chris Custer's answer for the extra digit by recalling the tangentially logarithmic "rule of $72$": An interest rate of $2.4$ percent will double the principle after $72/2.4 = 30$ periods.  So $(2^{10})^{30}$ will approximately equal $2 \times 10^{90}$, rather than just $10^{90}$.  That cuts the number of digits from $211$ to $210$.
A: 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.
A: Logarithm base 10 Scale will better answer your question.

Range                                    Digits

1-log(10)                                   1

log(10)-log(10^2)          2

log(10^2)-log(10^3)   3

log(10^3)-log(10^4)   4

log(10^n-1)-log(10^n)  n

I think you get the Idea.
Now for your number $$125^{100} = 5^{300}$$
$$\log_{10}{5^{300}} = 300 * \log_{10}{5} = 209.691$$
That means it is between 209 and 210. From the above table pattern, you can confirm there will be 210 digits.
A: For any natural number n the number of digits is $ 1+[log(n)]$ where $ [log(n)]$ stands for the integer part of $log(n)$
For $n=125^{100}$, 
 we get $$log(n)= 100 log(125)=209.6910013...$$
Thus the number of digits in $125^{100}$ is $210.$   
A: For all $x$ we have 
$$125^{100} < 125^{x}128^{100-x}=5^{3x}2^{700-7x}$$
Making $700-7x=3x$ we get $x=70$ and hence 
$$125^{100} < 5^{210}2^{210}=10^{210}$$
[I could had compared these two numbers directly, but using the $x$ above shows how I figured out that 210 is the key].
Since the approximation above is very close, intuition tells us that $125^{100}$ should have 210 digits. But intuiton is not enough, we need to prove it. Since we showed $125^{100} <10^{210}$, to conclude that 
$125^{100}$ has 210 digits we need to show
$$125^{100} \geq 10^{209}$
Lets prove it
$$125^{100} \geq 10^{209} \Leftrightarrow \\
5^{300} \geq 2^{209} 5^{209} \Leftrightarrow \\
5^{91} \geq 2^{209}  \Leftrightarrow \\
5^{13} \geq 2^{29.86}$$
The last inequality holds since $5^{13}>2^{30}$, but I don't see any simple way of getting this without a calculator.
