Question

Find the value of  $\log_{20} 100 + \log_{20} 1000 + \log_{20} 10000 \quad \bigl[\textit{Assume that } \log 2 = 0.3\bigr].$ 
 

• A. $90/13$
• B. $80/13$
• C. $110/13$
• D. $70/13$

Hint:

For $\log_a b$ with common logs, use $\log_a b=\frac{\log b}{\log a}$. Precompute $\log a$ once.

Answer 1

The Correct Option is A

Use change of base: $\log_{20}N=\dfrac{\log N}{\log 20}$. Since $\log 20=\log(2\cdot 10)=\log 2+1=1.3$, \[ \log_{20}100+\log_{20}1000+\log_{20}10000 =\frac{2+3+4}{1.3} =\frac{9}{1.3}=\frac{90}{13}. \] 

Answer 2

Instead of converting each logarithm separately and adding three fractions, first combine the three logarithms into one using the product rule, then convert just once.

\[ \log_{20}100+\log_{20}1000+\log_{20}10000=\log_{20}\left(100\times1000\times10000\right)=\log_{20}\left(10^{9}\right). \]

Since \( \log20=\log(2\times10)=\log2+\log10=0.3+1=1.3 \) (all base-\( 10 \)), and \( \log(10^9)=9 \), \[ \log_{20}(10^9)=\frac{9}{1.3}=\frac{90}{13}. \]

1. Option A (\( 90/13 \)): This matches the value computed directly above.
2. Option B (\( 80/13 \)): This would require the combined power of \( 10 \) to be \( 8 \) instead of \( 9 \), which does not match \( 100\times1000\times10000=10^9 \); rejected.
3. Option C (\( 110/13 \)): This would require a power of \( 11 \), inconsistent with the product \( 10^9 \); rejected.
4. Option D (\( 70/13 \)): This would require a power of \( 7 \), again inconsistent with \( 10^9 \); rejected.

Combining the three logarithms into a single term before converting confirms the value is \( 90/13 \).

Hence, the correct answer is \( 90/13 \).