Solve for I 148=10 log of I/(10^-12)

$148 = 10 \log (\frac{I}{10^{- 12}})$

Step 1

Rewrite the equation as $10 \log (\frac{I}{10^{- 12}}) = 148$ .

$10 \log (\frac{I}{10^{- 12}}) = 148$

Step 2

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

$10 \log (I \cdot 10^{12}) = 148$

Step 3

Divide each term in $10 \log (I \cdot 10^{12}) = 148$ by $10$ and simplify.

Divide each term in $10 \log (I \cdot 10^{12}) = 148$ by $10$ .

$\frac{10 \log (I \cdot 10^{12})}{10} = \frac{148}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 \log (I \cdot 10^{12})}{10} = \frac{148}{10}$

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

$\log (I \cdot 10^{12}) = \frac{148}{10}$

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

$\log (I) + \log (10^{12}) = \frac{148}{10}$

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

$\log (I) + 12 \log (10) = \frac{148}{10}$

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

$\log (I) + 12 \cdot 1 = \frac{148}{10}$

Multiply $12$ by $1$ .

$\log (I) + 12 = \frac{148}{10}$

Simplify the right side.

Cancel the common factor of $148$ and $10$ .

Factor $2$ out of $148$ .

$\log (I) + 12 = \frac{2 (74)}{10}$

Cancel the common factors.

Factor $2$ out of $10$ .

$\log (I) + 12 = \frac{2 \cdot 74}{2 \cdot 5}$

Cancel the common factor.

$\log (I) + 12 = \frac{2 \cdot 74}{2 \cdot 5}$

Rewrite the expression.

$\log (I) + 12 = \frac{74}{5}$

Step 4

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

Subtract $12$ from both sides of the equation.

$\log (I) = \frac{74}{5} - 12$

To write $- 12$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\log (I) = \frac{74}{5} - 12 \cdot \frac{5}{5}$

Combine $- 12$ and $\frac{5}{5}$ .

$\log (I) = \frac{74}{5} + \frac{- 12 \cdot 5}{5}$

Combine the numerators over the common denominator.

$\log (I) = \frac{74 - 12 \cdot 5}{5}$

Simplify the numerator.

Multiply $- 12$ by $5$ .

$\log (I) = \frac{74 - 60}{5}$

Subtract $60$ from $74$ .

$\log (I) = \frac{14}{5}$

Step 5

Rewrite $\log (I) = \frac{14}{5}$ 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 $b^{y} = x$ .

$10^{\frac{14}{5}} = I$

Step 6

Rewrite the equation as $I = 10^{\frac{14}{5}}$ .

$I = 10^{\frac{14}{5}}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$I = 10^{\frac{14}{5}}$

Decimal Form:

$I = 630.95734448 \dots$

Solve for I 148=10 log of I/(10^-12)

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