#### Answer

$I=I_0\left(10^{\frac{d}{10}}\right)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
d=10\log \left( \dfrac{I}{I_0} \right)
,$ in terms of $
I
,$ use the properties of equality to isolate the logarithmic expression. Then change to exponential form. Finally use again the properties of equality to isolate the needed variable.
$\bf{\text{Solution Details:}}$
Using the properties of equality, 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 to isolate the needed variable results to
\begin{array}{l}\require{cancel}
I_0\cdot10^{\frac{d}{10}}=I_0\cdot\dfrac{I}{I_0}
\\\\
I_0\left(10^{\frac{d}{10}}\right)=I
\\\\
I=I_0\left(10^{\frac{d}{10}}\right)
.\end{array}