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

Expand $\ln ({(\frac{4}{3})}^{1 - x})$ by moving $1 - x$ outside the logarithm.

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

Step 3

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

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

Step 4

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

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

Step 5

Solve the equation for $x$ .

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Step 5.1

Simplify $(1 - x) (\ln (4) - \ln (3))$ .

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Step 5.1.1

Rewrite.

$0 + 0 + (1 - x) (\ln (4) - \ln (3)) = x \ln (5)$

Step 5.1.2

Simplify by adding zeros.

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

Step 5.1.3

Expand $(1 - x) (\ln (4) - \ln (3))$ using the FOIL Method.

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Step 5.1.3.1

Apply the distributive property.

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

Step 5.1.3.2

Apply the distributive property.

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

Step 5.1.3.3

Apply the distributive property.

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

Step 5.1.4

Simplify each term.

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Step 5.1.4.1

Multiply $\ln (4)$ by $1$ .

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

Step 5.1.4.2

Multiply $- \ln (3)$ by $1$ .

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

Step 5.1.4.3

Rewrite using the commutative property of multiplication.

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

Step 5.1.4.4

Multiply $- 1$ by $- 1$ .

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

Step 5.1.4.5

Multiply $x$ by $1$ .

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

Step 5.2

Subtract $x \ln (5)$ from both sides of the equation.

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

Step 5.3

Move all terms not containing $x$ to the right side of the equation.

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Step 5.3.1

Subtract $\ln (4)$ from both sides of the equation.

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

Step 5.3.2

Add $\ln (3)$ to both sides of the equation.

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

Step 5.4

Factor $x$ out of $- x \ln (4) + x \ln (3) - x \ln (5)$ .

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Step 5.4.1

Factor $x$ out of $- x \ln (4)$ .

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

Step 5.4.2

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

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

Step 5.4.3

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

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

Step 5.4.4

Factor $x$ out of $x (- 1 \ln (4)) + x (\ln (3))$ .

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

Step 5.4.5

Factor $x$ out of $x (- 1 \ln (4) + \ln (3)) + x (- 1 \ln (5))$ .

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

Step 5.5

Rewrite $- 1 \ln (4)$ as $- \ln (4)$ .

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

Step 5.6

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

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

Step 5.7

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

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Step 5.7.1

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

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

Step 5.7.2

Simplify the left side.

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Step 5.7.2.1

Cancel the common factor of $- \ln (4) + \ln (3) - \ln (5)$ .

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Step 5.7.2.1.1

Cancel the common factor.

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

Step 5.7.2.1.2

Divide $x$ by $1$ .

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

Step 5.7.3

Simplify the right side.

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Step 5.7.3.1

Combine the numerators over the common denominator.

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

Step 5.7.3.2

Factor $- 1$ out of $- \ln (4)$ .

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

Step 5.7.3.3

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

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

Step 5.7.3.4

Factor $- 1$ out of $- \ln (4) - 1 (- \ln (3))$ .

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

Step 5.7.3.5

Rewrite $- (\ln (4) - \ln (3))$ as $- 1 (\ln (4) - \ln (3))$ .

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

Step 5.7.3.6

Factor $- 1$ out of $- \ln (4)$ .

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

Step 5.7.3.7

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

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

Step 5.7.3.8

Factor $- 1$ out of $- \ln (4) - 1 (- \ln (3))$ .

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

Step 5.7.3.9

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

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

Step 5.7.3.10

Factor $- 1$ out of $- (\ln (4) - \ln (3)) - \ln (5)$ .

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

Step 5.7.3.11

Rewrite $- (\ln (4) - \ln (3) + \ln (5))$ as $- 1 (\ln (4) - \ln (3) + \ln (5))$ .

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

Step 5.7.3.12

Cancel the common factor.

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

Step 5.7.3.13

Rewrite the expression.

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.15164147 \dots$