What is the value of \frac{\log_9 8 \cdot \log_4 25}{\log_3 5}? — JAMB UTME

What is the value of $\frac{\log_9 8 \cdot \log_4 25}{\log_3 5}$ ?

\frac{3}{2}

Applying the power identity

\log_{b^k} a^m = \frac{m}{k} \log_b a

yields

\log_9 8 = \frac{3}{2} \log_3 2

and

\log_4 25 = \log_2 5

. Using change of base,

\log_3 2 \cdot \log_2 5 = \log_3 5

. Thus, the numerator simplifies to

\frac{3}{2} \log_3 5

. Dividing by

\log_3 5

leaves

\frac{3}{2}

Rewrite logarithmic expressions with base powers in terms of prime bases

\log_9 8 = \log_{3^2} (2^3) = \frac{3}{2} \log_3 2

and

\log_4 25 = \log_{2^2} (5^2) = \frac{2}{2} \log_2 5 = \log_2 5

Applying the logarithmic identity

\log_{b^k} (a^m) = \frac{m}{k} \log_b a

simplifies base powers.

Multiply the numerator terms together using the change of base rule

(\log_9 8)(\log_4 25) = \left(\frac{3}{2} \log_3 2\right) \cdot (\log_2 5) = \frac{3}{2} (\log_3 2 \cdot \log_2 5) = \frac{3}{2} \log_3 5$

By change of base,

\log_3 2 \cdot \log_2 5 = \frac{\log 2}{\log 3} \cdot \frac{\log 5}{\log 2} = \frac{\log 5}{\log 3} = \log_3 5

Divide the numerator result by the denominator

\frac{\frac{3}{2} \log_3 5}{\log_3 5} = \frac{3}{2}

The term

\log_3 5

cancels out from numerator and denominator.

Change of Base and Exponent Rules for Logarithms

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