WAEC - Further Mathematics (2018 - No. 8)
If \(\log_{3}a - 2 = 3\log_{3}b\), express a in terms of b.
\(a = b^{3} - 3\)
\(a = b^{3} - 9\)
\(a = 9b^{3}\)
\(a = \frac{b^{3}}{9}\)
Explanation
\(\log_{3}a - 2 = 3\log_{3}b\)
Using the laws of logarithm, we know that \( 2 = 2\log_{3}3 = \log_{3}3^{2}\)
\(\therefore \log_{3}a - \log_{3}3^{2} = \log_{3}b^{3}\)
= \(\log_{3}(\frac{a}{3^{2}}) = \log_{3}b^{3} \implies \frac{a}{9} = b^{3}\)
\(\implies a = 9b^{3}\)
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