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I was given a log equation:

$$D = 10 \log (I/I_0) $$

$I$ is the unknown in this case, $I_0 = 10^{-12}$ and $D = 89.3$.

I did the following steps:

$$ \begin{aligned} \ 89.3 &= 10 \log \left(\frac{I}{10^{-12}}\right) \\ \ \frac{89.3}{10} &= \log \left(\frac{I}{10^{-12}}\right) \\ \ 8.93 &= \log \left(\frac{I}{10^{-12}}\right) \\ \end{aligned} $$

I'm not quite sure how to isolate I after step 3, and I'm also unsure if dividing $89.3/10$ is correct as well. So how can I find the unknown ($I$)?

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Yes you are on the right track. For logarithms we have $$\log_a y=x\iff a^x=y$$ Here's a tip: $\log\left(\frac{I}{10^{-12}}\right)=\log(10^{12}I).$ So then we have $$8.93=\log_{10}(10^{12}I)\implies 10^{8.93}=10^{12}I\implies 10^{8.93-12}=I\implies I=10^{-3.07}$$

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You're on the right way. You need to find the function f so that $f(log(x))=x$
Hint
$\forall x \in \mathbb{R}, log_a(a^x)=x, a\in \mathbb{R}$

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