Answer
(a) $-3$
(b) $\dfrac{1}{2}$
(c) $-1$
Work Step by Step
RECALL:
(1) $\log_b{b}=1$
(2) $\log_b{1}=0$
(3) $\log_b{b^x}=x$
(a)
$\frac{1}{27}$ is equivalent to $\frac{1}{3^3}$ so the given expression can be written as:
$=\log_3{(\frac{1}{3^3})}$
Since $\dfrac{1}{a^m} = a^{-m}$, the given expression is equivalent to:
$=\log_3{(3^{-3})}$
Use rule (3) above to obtain:
$=-3$
(b)
RECALL: $\sqrt{a} = a^{\frac{1}{2}}$
Thus, the given expression is equivalent to:
$=\log_{10}{(10^{\frac{1}{2}})}$
Use rule (3) above to obtain:
$=\dfrac{1}{2}$
(c)
Note that $0.2 = \dfrac{1}{5}$.
Use the rule $\dfrac{1}{a^m} = a^{-m}$ to obtain:
$\dfrac{1}{5} = 5^{-1}$
This means that the given expression is equivalent to:
$=\log_5{(5^{-1})}$
Use rule (3) above to obtain:
$=-1$
