{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
{{ searchError }}
search
{{ courseTrack.displayTitle }}
To evaluate the given logarithm, we can rewrite it as an exponential equation by using the definition of a logarithm. $\begin{gathered} \log_{{\color{#0000FF}{b}}}({\color{#FF0000}{x}})=\textcolor{purple}{y} \quad \Leftrightarrow \quad {\color{#FF0000}{x}}={\color{#0000FF}{b}}^{\textcolor{purple}{y}} \end{gathered}$ The above means that the logarithm $\textcolor{purple}{y}$ is the exponent to which ${\color{#0000FF}{b}}$ must be raised to get ${\color{#FF0000}{x}}.$ For our exercise, $\textcolor{purple}{y}$ is the exponent to which ${\color{#0000FF}{8}}$ must be raised to get ${\color{#FF0000}{\frac{1}{512}}}.$ $\begin{gathered} \log_{{\color{#0000FF}{8}}}\left({\color{#FF0000}{\frac{1}{512}}}\right)=\textcolor{purple}{y} \quad \Leftrightarrow \quad {\color{#FF0000}{\frac{1}{512}}}={\color{#0000FF}{8}}^{\textcolor{purple}{y}} \end{gathered}$ Therefore, we should rewrite $\frac{1}{512}$ as a power of $8$ and the resulting exponent will be our answer.
$\dfrac{1}{512}$
$\dfrac{1}{a}=a^{\text{-} 1}$
$512^{\text{-}1}$
$\left(8^3\right)^{\text{-}1}$
$8^{\text{-}3}$
Thus, $\log_8 \left(\frac{1}{512}\right)=\text{-}3.$