Precalculus Examples

$\log (S) = \log (c) + k \log (A)$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

$\log (S) - \log (c) - k \log (A) = 0$

Step 2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log (\frac{S}{c}) - k \log (A) = 0$

Step 3

Add $k \log (A)$ to both sides of the equation.

$\log (\frac{S}{c}) = k \log (A)$

Step 4

Rewrite $\log (\frac{S}{c}) = k \log (A)$ 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 $b^{y} = x$ .

$10^{k \log (A)} = \frac{S}{c}$

Step 5

Solve for $S$ .

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Step 5.1

Rewrite the equation as $\frac{S}{c} = 10^{k \log (A)}$ .

$\frac{S}{c} = 10^{k \log (A)}$

Step 5.2

Multiply both sides by $c$ .

$\frac{S}{c} c = 10^{k \log (A)} c$

Step 5.3

Simplify.

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Step 5.3.1

Simplify the left side.

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Step 5.3.1.1

Cancel the common factor of $c$ .

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Step 5.3.1.1.1

Cancel the common factor.

$\frac{S}{c} c = 10^{k \log (A)} c$

Step 5.3.1.1.2

Rewrite the expression.

$S = 10^{k \log (A)} c$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Reorder factors in $10^{k \log (A)} c$ .

$S = c \cdot 10^{k \log (A)}$