JAMB 1991 · UME · Q9

Simplify $2\log\frac25 - \log\frac{72}{125} + \log 9$ .

Worked solution (try it first)

Move the 2 up as a power: $2\log\frac25 = \log\frac{4}{25}$ .
Combine the logs: $\log\left(\frac{4}{25} \times \frac{125}{72} \times 9\right)$ .
The number inside is $\frac{4500}{1800} = \frac52$ .
$\frac52 = \frac{10}{4}$ , so $\log\frac52 = \log 10 - \log 4$ .
With $\log 10 = 1$ and $\log 4 = 2\log 2$ , the value is $1 - 2\log 2$ , option D.