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Evaluating Logarithms

Evaluating Logarithms 1.6 - Solution

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To evaluate the given logarithm, we can rewrite it as an exponential equation by using the definition of a logarithm. logb(x)=yx=by\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 y\textcolor{purple}{y} is the exponent to which b{\color{#0000FF}{b}} must be raised to get x.{\color{#FF0000}{x}}. For our exercise, y\textcolor{purple}{y} is the exponent to which 8{\color{#0000FF}{8}} must be raised to get 1512.{\color{#FF0000}{\frac{1}{512}}}. log8(1512)=y1512=8y\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 1512\frac{1}{512} as a power of 88 and the resulting exponent will be our answer.
1512\dfrac{1}{512}
1a=a-1\dfrac{1}{a}=a^{\text{-} 1}
512-1512^{\text{-}1}
(83)-1\left(8^3\right)^{\text{-}1}
8-38^{\text{-}3}
Thus, log8(1512)=-3.\log_8 \left(\frac{1}{512}\right)=\text{-}3.