Question

How do you solve for 'I' in these cases?

10 log (I/10^-12) = 120
and
10 log (I/10) = 0

Thanks a lot for your time!

Answer 1

@freckles
@pooja195
@Agl202

Answer 2

@IrishBoy123

Answer 3

is it \[10 log (\dfrac{I}{10^{-12}}) = 120\]?

and \(\log_{10}\)??

Answer 4

10 log (I/10^-12)

Answer 5

and yes to the first one

Answer 6

@IrishBoy123

Answer 7

@Michele_Laino

Answer 8

hint:
we have these steps:
\[\Large \begin{gathered}
\log \left( {\frac{I}{{{{10}^{ - 12}}}}} \right) = \frac{{120}}{{10}} \hfill \\
\hfill \\
\log \left( {\frac{I}{{{{10}^{ - 12}}}}} \right) = 12 \hfill \\
\end{gathered} \]

Answer 9

so, if we apply the definition of logarithm, we can write this:
\[\huge \frac{I}{{{{10}^{ - 12}}}} = {10^{12}}\]

Answer 10

hint:
if I multiply both sides by \(10^{-12}\), I get:

\[\huge \frac{I}{{{{10}^{ - 12}}}} \times {10^{ - 12}} = {10^{12}} \times {10^{ - 12}}\]
please continue