Question

Express $\log 2^3-\log 64+\frac{1}{2}\log 16$ as a single logarithm.

Answer 1

We have,

$=\log 2^3-\log 64+\frac{1}{2}\log 16$

$=\log 8-\log 64+\log \sqrt{16}$ $\because n\log a=\log a^n$

$=\log 8-\log 64+\log 4$

$=\log 32-\log 64$ $\because \log m+\log n=\log mn$

$=\log \frac{1}{2}$

Hence, this is the answer.