How many figures do they have?

Question:
Knowing that $\log_{10}2=0.3010$ and $\log_{10}3=0.4771$,
(1) find the number of figures of $6^{50}$.
(2) find at which decimal place appears to be the first non-zero of $\left(\frac{1}{2}\right)^{300}$.

Answer:
(1) Take the log of $6^{50}$.
\[ \log_{10}6^{50} = 50\log_{10}(2\times 3)   \\
    = 50(\log_{10}2 + \log_{10}3) \\
    =  50(0.3010 + 0.4771)  \\
    = 38.905   \]
Thus, we can say
\[ 38<\log_{10}6^{50}<39 \]
Namely,
\[ 10^{38}<6^{50}<10^{39} \]
$6^{50}$ has 39 figures.

(2) Take the log.
\[ \log_{10}\left(\frac{1}{2}\right)^{300} = 300\log_{10}\frac{1}{2}  \\
      = 300(\log_{10}1 - \log_{10}2)  \\
      = 300(0-\log_{10}2)  \\
     = -90.3  \]
Then, we have
\[ -91<\log_{10}\left(\frac{1}{2}\right)^{300}< -90   \\
    10^{-91} < \left(\frac{1}{2}\right)^{300} < 10^{-90} \]
$\left(\frac{1}{2}\right)^{300}$ has non-zero number at the 91-st decimal place.

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