Solve for I 80=10 log of I/(10^-12)

$80 = 10 \log (\frac{I}{10^{- 12}})$

Step 1

Rewrite the equation as $10 \log (\frac{I}{10^{- 12}}) = 80$ .

$10 \log (\frac{I}{10^{- 12}}) = 80$

Step 2

Move $10^{- 12}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$10 \log (I \cdot 10^{12}) = 80$

Step 3

Divide each term in $10 \log (I \cdot 10^{12}) = 80$ by $10$ and simplify.

Divide each term in $10 \log (I \cdot 10^{12}) = 80$ by $10$ .

$\frac{10 \log (I \cdot 10^{12})}{10} = \frac{80}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 \log (I \cdot 10^{12})}{10} = \frac{80}{10}$

Divide $\log (I \cdot 10^{12})$ by $1$ .

$\log (I \cdot 10^{12}) = \frac{80}{10}$

Rewrite $\log (I \cdot 10^{12})$ as $\log (I) + \log (10^{12})$ .

$\log (I) + \log (10^{12}) = \frac{80}{10}$

Use logarithm rules to move $12$ out of the exponent.

$\log (I) + 12 \log (10) = \frac{80}{10}$

Logarithm base $10$ of $10$ is $1$ .

$\log (I) + 12 \cdot 1 = \frac{80}{10}$

Multiply $12$ by $1$ .

$\log (I) + 12 = \frac{80}{10}$

Simplify the right side.

Divide $80$ by $10$ .

$\log (I) + 12 = 8$

Step 4

Move all terms not containing $\log (I)$ to the right side of the equation.

Subtract $12$ from both sides of the equation.

$\log (I) = 8 - 12$

Subtract $12$ from $8$ .

$\log (I) = - 4$

Step 5

Rewrite $\log (I) = - 4$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$10^{- 4} = I$

Step 6

Solve for $I$ .

Rewrite the equation as $I = 10^{- 4}$ .

$I = 10^{- 4}$

Simplify $10^{- 4}$ .

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$I = \frac{1}{10^{4}}$

Raise $10$ to the power of $4$ .

$I = \frac{1}{10000}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$I = \frac{1}{10000}$

Decimal Form:

$I = 0.0001$

Solve for I 80=10 log of I/(10^-12)

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