Sure, let's solve the equation step-by-step.
We start with the given equation:
[tex]\[ 125 = 10 \log \left( \frac{I}{10^{-16}} \right) \][/tex]
1. Isolate the logarithm:
First, divide both sides by 10 to get the logarithm term by itself:
[tex]\[ \frac{125}{10} = \log \left( \frac{I}{10^{-16}} \right) \][/tex]
Simplifying the left-hand side:
[tex]\[ 12.5 = \log \left( \frac{I}{10^{-16}} \right) \][/tex]
2. Rewrite the equation in exponential form using properties of logarithms:
Recall that [tex]\( \log_b (x) = y \)[/tex] means [tex]\( x = b^y \)[/tex]. Here, we're using a common logarithm (base 10):
[tex]\[ 10^{12.5} = \frac{I}{10^{-16}} \][/tex]
3. Isolate [tex]\( I \)[/tex]:
To solve for [tex]\( I \)[/tex], multiply both sides of the equation by [tex]\( 10^{-16} \)[/tex]:
[tex]\[ I = 10^{12.5} \times 10^{-16} \][/tex]
4. Simplify the expression:
Using the properties of exponents, specifically [tex]\( 10^a \times 10^b = 10^{a+b} \)[/tex]:
[tex]\[ I = 10^{12.5 - 16} \][/tex]
Simplify the exponent:
[tex]\[ I = 10^{-3.5} \][/tex]
5. Convert the result to a decimal form:
To find [tex]\( I \)[/tex] in decimal form, we need to evaluate [tex]\( 10^{-3.5} \)[/tex]:
[tex]\[ 10^{-3.5} \approx 0.00031622776601683794 \][/tex]
Thus, the value of [tex]\( I \)[/tex] is approximately [tex]\( 0.00031622776601683794 \)[/tex].
