How many digits are in $125^{100}$?

$2^{10}\approx 10^3$, so approximately, $\frac {1000^{100}}{10^{90}} =\frac {100^{100}\cdot 10^{100}}{10^{90}}=100^{100}\cdot 10^{10}=10^{210}$... So about $211$.

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 $= [209.70]+1 = 210.$

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 $210$.