Precalculus Examples

${(\frac{9}{4})}^{x} = 5^{1 - x}$

Step 1

Take the log of both sides of the equation.

$\ln ({(\frac{9}{4})}^{x}) = \ln (5^{1 - x})$

Step 2

Expand $\ln ({(\frac{9}{4})}^{x})$ by moving $x$ outside the logarithm.

$x \ln (\frac{9}{4}) = \ln (5^{1 - x})$

Step 3

Rewrite $\ln (\frac{9}{4})$ as $\ln (9) - \ln (4)$ .

$x (\ln (9) - \ln (4)) = \ln (5^{1 - x})$

Step 4

Rewrite $\ln (4)$ as $\ln (2^{2})$ .

$x (\ln (9) - \ln (2^{2})) = \ln (5^{1 - x})$

Step 5

Expand $\ln (2^{2})$ by moving $2$ outside the logarithm.

$x (\ln (9) - (2 \ln (2))) = \ln (5^{1 - x})$

Step 6

Multiply $2$ by $- 1$ .

$x (\ln (9) - 2 \ln (2)) = \ln (5^{1 - x})$

Step 7

Expand $\ln (5^{1 - x})$ by moving $1 - x$ outside the logarithm.

$x (\ln (9) - 2 \ln (2)) = (1 - x) \ln (5)$

Step 8

Solve the equation for $x$ .

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Step 8.1

Simplify $x (\ln (9) - 2 \ln (2))$ .

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Step 8.1.1

Rewrite.

$0 + 0 + x (\ln (9) - 2 \ln (2)) = (1 - x) \ln (5)$

Step 8.1.2

Simplify by adding zeros.

$x (\ln (9) - 2 \ln (2)) = (1 - x) \ln (5)$

Step 8.1.3

Apply the distributive property.

$x \ln (9) + x (- 2 \ln (2)) = (1 - x) \ln (5)$

Step 8.1.4

Rewrite using the commutative property of multiplication.

$x \ln (9) - 2 x \ln (2) = (1 - x) \ln (5)$

Step 8.2

Simplify $(1 - x) \ln (5)$ .

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Step 8.2.1

Apply the distributive property.

$x \ln (9) - 2 x \ln (2) = 1 \ln (5) - x \ln (5)$

Step 8.2.2

Multiply $\ln (5)$ by $1$ .

$x \ln (9) - 2 x \ln (2) = \ln (5) - x \ln (5)$

Step 8.3

Add $x \ln (5)$ to both sides of the equation.

$x \ln (9) - 2 x \ln (2) + x \ln (5) = \ln (5)$

Step 8.4

Factor $x$ out of $x \ln (9) - 2 x \ln (2) + x \ln (5)$ .

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Step 8.4.1

Factor $x$ out of $x \ln (9)$ .

$x (\ln (9)) - 2 x \ln (2) + x \ln (5) = \ln (5)$

Step 8.4.2

Factor $x$ out of $- 2 x \ln (2)$ .

$x (\ln (9)) + x (- 2 \ln (2)) + x \ln (5) = \ln (5)$

Step 8.4.3

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

$x (\ln (9)) + x (- 2 \ln (2)) + x (\ln (5)) = \ln (5)$

Step 8.4.4

Factor $x$ out of $x (\ln (9)) + x (- 2 \ln (2))$ .

$x (\ln (9) - 2 \ln (2)) + x (\ln (5)) = \ln (5)$

Step 8.4.5

Factor $x$ out of $x (\ln (9) - 2 \ln (2)) + x (\ln (5))$ .

$x (\ln (9) - 2 \ln (2) + \ln (5)) = \ln (5)$

Step 8.5

Divide each term in $x (\ln (9) - 2 \ln (2) + \ln (5)) = \ln (5)$ by $\ln (9) - 2 \ln (2) + \ln (5)$ and simplify.

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Step 8.5.1

Divide each term in $x (\ln (9) - 2 \ln (2) + \ln (5)) = \ln (5)$ by $\ln (9) - 2 \ln (2) + \ln (5)$ .

$\frac{x (\ln (9) - 2 \ln (2) + \ln (5))}{\ln (9) - 2 \ln (2) + \ln (5)} = \frac{\ln (5)}{\ln (9) - 2 \ln (2) + \ln (5)}$

Step 8.5.2

Simplify the left side.

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Step 8.5.2.1

Cancel the common factor of $\ln (9) - 2 \ln (2) + \ln (5)$ .

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Step 8.5.2.1.1

Cancel the common factor.

$\frac{x (\ln (9) - 2 \ln (2) + \ln (5))}{\ln (9) - 2 \ln (2) + \ln (5)} = \frac{\ln (5)}{\ln (9) - 2 \ln (2) + \ln (5)}$

Step 8.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (5)}{\ln (9) - 2 \ln (2) + \ln (5)}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (5)}{\ln (9) - 2 \ln (2) + \ln (5)}$

Decimal Form:

$x = 0.66495583 \dots$