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) First, 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 state

\[ 38<\log_{10}6^{50}<39 \]

Namely,

\[ 10^{38}<6^{50}<10^{39} \]

\(6^{50}\) has 39 figures.

(2) Take the log of the given number.

\[ \log_{10}\left(\frac{1}{2}\right)^{300} = 300\log_{10}\frac{1}{2}  \]

\[      = 300(\log_{10}1 – \log_{10}2)  \]

\[      = 300(0-\log_{10}2)  \]

\[     = -90.3  \]

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.