Precalculus Examples

$\frac{4}{3^{1 - x}} = 5^{x}$

Step 1

Take the log of both sides of the equation.

$\ln (\frac{4}{3^{1 - x}}) = \ln (5^{x})$

Step 2

Rewrite $\ln (\frac{4}{3^{1 - x}})$ as $\ln (4) - \ln (3^{1 - x})$ .

$\ln (4) - \ln (3^{1 - x}) = \ln (5^{x})$

Step 3

Expand $\ln (3^{1 - x})$ by moving $1 - x$ outside the logarithm.

$\ln (4) - ((1 - x) \ln (3)) = \ln (5^{x})$

Step 4

Remove parentheses.

$\ln (4) - (1 - x) \ln (3) = \ln (5^{x})$

Step 5

Expand $\ln (5^{x})$ by moving $x$ outside the logarithm.

$\ln (4) - (1 - x) \ln (3) = x \ln (5)$

Step 6

Solve the equation for $x$ .

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Step 6.1

Simplify the left side.

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Step 6.1.1

Simplify each term.

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Step 6.1.1.1

Apply the distributive property.

$\ln (4) + (- 1 \cdot 1 + x) \ln (3) = x \ln (5)$

Step 6.1.1.2

Multiply $- 1$ by $1$ .

$\ln (4) + (- 1 + x) \ln (3) = x \ln (5)$

Step 6.1.1.3

Multiply $- - x$ .

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Step 6.1.1.3.1

Multiply $- 1$ by $- 1$ .

$\ln (4) + (- 1 + 1 x) \ln (3) = x \ln (5)$

Step 6.1.1.3.2

Multiply $x$ by $1$ .

$\ln (4) + (- 1 + x) \ln (3) = x \ln (5)$

Step 6.1.1.4

Apply the distributive property.

$\ln (4) - 1 \ln (3) + x \ln (3) = x \ln (5)$

Step 6.1.1.5

Rewrite $- 1 \ln (3)$ as $- \ln (3)$ .

$\ln (4) - \ln (3) + x \ln (3) = x \ln (5)$

Step 6.1.2

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

$\ln (\frac{4}{3}) + x \ln (3) = x \ln (5)$

Step 6.1.3

Reorder $\ln (\frac{4}{3})$ and $x \ln (3)$ .

$x \ln (3) + \ln (\frac{4}{3}) = x \ln (5)$

Step 6.2

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

$x \ln (3) + \ln (\frac{4}{3}) - x \ln (5) = 0$

Step 6.3

Subtract $\ln (\frac{4}{3})$ from both sides of the equation.

$x \ln (3) - x \ln (5) = - \ln (\frac{4}{3})$

Step 6.4

Factor $x$ out of $x \ln (3) - x \ln (5)$ .

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Step 6.4.1

Factor $x$ out of $x \ln (3)$ .

$x (\ln (3)) - x \ln (5) = - \ln (\frac{4}{3})$

Step 6.4.2

Factor $x$ out of $- x \ln (5)$ .

$x (\ln (3)) + x (- 1 \ln (5)) = - \ln (\frac{4}{3})$

Step 6.4.3

Factor $x$ out of $x (\ln (3)) + x (- 1 \ln (5))$ .

$x (\ln (3) - 1 \ln (5)) = - \ln (\frac{4}{3})$

Step 6.5

Rewrite $- 1 \ln (5)$ as $- \ln (5)$ .

$x (\ln (3) - \ln (5)) = - \ln (\frac{4}{3})$

Step 6.6

Divide each term in $x (\ln (3) - \ln (5)) = - \ln (\frac{4}{3})$ by $\ln (3) - \ln (5)$ and simplify.

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Step 6.6.1

Divide each term in $x (\ln (3) - \ln (5)) = - \ln (\frac{4}{3})$ by $\ln (3) - \ln (5)$ .

$\frac{x (\ln (3) - \ln (5))}{\ln (3) - \ln (5)} = \frac{- \ln (\frac{4}{3})}{\ln (3) - \ln (5)}$

Step 6.6.2

Simplify the left side.

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Step 6.6.2.1

Cancel the common factor of $\ln (3) - \ln (5)$ .

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Step 6.6.2.1.1

Cancel the common factor.

$\frac{x (\ln (3) - \ln (5))}{\ln (3) - \ln (5)} = \frac{- \ln (\frac{4}{3})}{\ln (3) - \ln (5)}$

Step 6.6.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (\frac{4}{3})}{\ln (3) - \ln (5)}$

Step 6.6.3

Simplify the right side.

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Step 6.6.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{4}{3})}{\ln (3) - \ln (5)}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{\ln (\frac{4}{3})}{\ln (3) - \ln (5)}$

Decimal Form:

$x = 0.56317079 \dots$