Question 251335
It took me a while to understand what was going on. The equation is
{{{L = 10log((I/I[0]))}}}
Those are i's in the logarithm, not 1's. And the zero in the logarithm is a subscript, not the digit of a number. I think this is where you got confused. (I know this is where I got confused when I first looked at the problem.)<br>
Since {{{I[0] = 10^(-12)}}}, your teacher's solution:
{{{L =10log((I/I[0]))}}}
{{{111 = 10log((I/10^(-12)))}}}
{{{11.1=log((I)) - log((10^(-12)))}}}
is correct so far.<br>
And, as your teacher says, we still need to solve for I. Solving for I means we have to get an equation we I is by itself on one side.<br>
We can start by finding the {{{log(( 10^(-12)))}}}/ If we understand logarithms we will know that this must be -12. If you don't see this:<ul><li>Ask your calculator</li><li>Use the property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to rewrite {{{log(( 10^(-12)))}}} as {{{-12*log(( 10))}}}. And since {{{log((10)) = 1}}} by definition this becomes -12.</li></ul>
Our equation is now:
{{{11.1=log((I)) - (-12)}}}
or
{{{11.1=log((I)) + 12}}}
Next we can subtract 12 from each side:
{{{-0.9 = log((I))}}}
If you know how to do inverse logarithms on your calculator you can find "I" now. If not, then we can rewrite this in exponential form:
{{{10^(-0.9) = I}}}
Now we can find "I" by using our calculators to raise 10 to the -0.9 power. (I'll leave this step up to you.)