College Algebra (11th Edition)

$I_0=\dfrac{I}{10^{\frac{d}{10}}}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $d=10\log \left( \dfrac{I}{I_0} \right) ,$ for $I_0 ,$ use the properties of equality to isolate the $\log$ expression. Then change to exponential form. Use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Dividing both sides by $10,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{d}{10}=\dfrac{10\log \left( \dfrac{I}{I_0} \right)}{10} \\\\ \dfrac{d}{10}=\log \left( \dfrac{I}{I_0} \right) .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} \dfrac{d}{10}=\log_{10} \left( \dfrac{I}{I_0} \right) \\\\ 10^{\frac{d}{10}}=\dfrac{I}{I_0} .\end{array} Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} I_0\left( 10^{\frac{d}{10}}\right)=I_0\cdot\dfrac{I}{I_0} \\\\ I_0\left( 10^{\frac{d}{10}}\right)=I \\\\ \dfrac{I_0\left( 10^{\frac{d}{10}}\right)}{10^{\frac{d}{10}}}=\dfrac{I}{10^{\frac{d}{10}}} \\\\ I_0=\dfrac{I}{10^{\frac{d}{10}}} .\end{array}