Precalculus Examples

${Pa}^{x} = Q b^{x}$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln ({Pa}^{x}) = \ln (Q b^{x})$

Step 2

Expand $\ln ({Pa}^{x})$ by moving $x$ outside the logarithm.

$x \ln (Pa) = \ln (Q b^{x})$

Step 3

Expand the right side.

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Step 3.1

Rewrite $\ln (Q b^{x})$ as $\ln (Q) + \ln (b^{x})$ .

$x \ln (Pa) = \ln (Q) + \ln (b^{x})$

Step 3.2

Expand $\ln (b^{x})$ by moving $x$ outside the logarithm.

$x \ln (Pa) = \ln (Q) + x \ln (b)$

Step 4

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

$x \ln (Pa) - \ln (Q) - x \ln (b) = 0$

Step 5

Add $\ln (Q)$ to both sides of the equation.

$x \ln (Pa) - x \ln (b) = \ln (Q)$

Step 6

Factor $x$ out of $x \ln (Pa) - x \ln (b)$ .

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Step 6.1

Factor $x$ out of $- x \ln (b)$ .

$x \ln (Pa) + x (- 1 \ln (b)) = \ln (Q)$

Step 6.2

Factor $x$ out of $x \ln (Pa) + x (- 1 \ln (b))$ .

$x (\ln (Pa) - 1 \ln (b)) = \ln (Q)$

Step 7

Rewrite $- 1 \ln (b)$ as $- \ln (b)$ .

$x (\ln (Pa) - \ln (b)) = \ln (Q)$

Step 8

Divide each term in $x (\ln (Pa) - \ln (b)) = \ln (Q)$ by $\ln (Pa) - \ln (b)$ and simplify.

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Step 8.1

Divide each term in $x (\ln (Pa) - \ln (b)) = \ln (Q)$ by $\ln (Pa) - \ln (b)$ .

$\frac{x (\ln (Pa) - \ln (b))}{\ln (Pa) - \ln (b)} = \frac{\ln (Q)}{\ln (Pa) - \ln (b)}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $\ln (Pa) - \ln (b)$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{x (\ln (Pa) - \ln (b))}{\ln (Pa) - \ln (b)} = \frac{\ln (Q)}{\ln (Pa) - \ln (b)}$

Step 8.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (Q)}{\ln (Pa) - \ln (b)}$

Step 9

Simplify the denominator.

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Step 9.1

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

$x = \frac{\ln (Q)}{\ln (\frac{Pa}{b})}$

Step 9.2

Rewrite the expression.

$x = \frac{\ln (Q)}{\ln (\frac{1}{b} Pa)}$