Precalculus Examples

$2^{1 - x} = 3^{x}$

Step 1

Take the log of both sides of the equation.

$\ln (2^{1 - x}) = \ln (3^{x})$

Step 2

Expand $\ln (2^{1 - x})$ by moving $1 - x$ outside the logarithm.

$(1 - x) \ln (2) = \ln (3^{x})$

Step 3

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

$(1 - x) \ln (2) = x \ln (3)$

Step 4

Solve the equation for $x$ .

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Step 4.1

Simplify $(1 - x) \ln (2)$ .

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Step 4.1.1

Rewrite.

$0 + 0 + (1 - x) \ln (2) = x \ln (3)$

Step 4.1.2

Simplify by adding zeros.

$(1 - x) \ln (2) = x \ln (3)$

Step 4.1.3

Apply the distributive property.

$1 \ln (2) - x \ln (2) = x \ln (3)$

Step 4.1.4

Multiply $\ln (2)$ by $1$ .

$\ln (2) - x \ln (2) = x \ln (3)$

Step 4.2

Subtract $x \ln (3)$ from both sides of the equation.

$\ln (2) - x \ln (2) - x \ln (3) = 0$

Step 4.3

Subtract $\ln (2)$ from both sides of the equation.

$- x \ln (2) - x \ln (3) = - \ln (2)$

Step 4.4

Factor $x$ out of $- x \ln (2) - x \ln (3)$ .

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Step 4.4.1

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

$x (- 1 \ln (2)) - x \ln (3) = - \ln (2)$

Step 4.4.2

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

$x (- 1 \ln (2)) + x (- 1 \ln (3)) = - \ln (2)$

Step 4.4.3

Factor $x$ out of $x (- 1 \ln (2)) + x (- 1 \ln (3))$ .

$x (- 1 \ln (2) - 1 \ln (3)) = - \ln (2)$

Step 4.5

Rewrite $- 1 \ln (2)$ as $- \ln (2)$ .

$x (- \ln (2) - 1 \ln (3)) = - \ln (2)$

Step 4.6

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

$x (- \ln (2) - \ln (3)) = - \ln (2)$

Step 4.7

Divide each term in $x (- \ln (2) - \ln (3)) = - \ln (2)$ by $- \ln (2) - \ln (3)$ and simplify.

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Step 4.7.1

Divide each term in $x (- \ln (2) - \ln (3)) = - \ln (2)$ by $- \ln (2) - \ln (3)$ .

$\frac{x (- \ln (2) - \ln (3))}{- \ln (2) - \ln (3)} = \frac{- \ln (2)}{- \ln (2) - \ln (3)}$

Step 4.7.2

Simplify the left side.

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Step 4.7.2.1

Cancel the common factor of $- \ln (2) - \ln (3)$ .

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Step 4.7.2.1.1

Cancel the common factor.

$\frac{x (- \ln (2) - \ln (3))}{- \ln (2) - \ln (3)} = \frac{- \ln (2)}{- \ln (2) - \ln (3)}$

Step 4.7.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (2)}{- \ln (2) - \ln (3)}$

Step 4.7.3

Simplify the right side.

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Step 4.7.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (2)}{- \ln (2) - \ln (3)}$

Step 4.7.3.2

Factor $- 1$ out of $- \ln (2)$ .

$x = - \frac{\ln (2)}{- \ln (2) - \ln (3)}$

Step 4.7.3.3

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

$x = - \frac{\ln (2)}{- \ln (2) - \ln (3)}$

Step 4.7.3.4

Factor $- 1$ out of $- \ln (2) - \ln (3)$ .

$x = - \frac{\ln (2)}{- (\ln (2) + \ln (3))}$

Step 4.7.3.5

Simplify the expression.

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Step 4.7.3.5.1

Rewrite $- (\ln (2) + \ln (3))$ as $- 1 (\ln (2) + \ln (3))$ .

$x = - \frac{\ln (2)}{- 1 (\ln (2) + \ln (3))}$

Step 4.7.3.5.2

Move the negative in front of the fraction.

$x = - - \frac{\ln (2)}{\ln (2) + \ln (3)}$

Step 4.7.3.5.3

Multiply $- 1$ by $- 1$ .

$x = 1 \frac{\ln (2)}{\ln (2) + \ln (3)}$

Step 4.7.3.5.4

Multiply $\frac{\ln (2)}{\ln (2) + \ln (3)}$ by $1$ .

$x = \frac{\ln (2)}{\ln (2) + \ln (3)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (2)}{\ln (2) + \ln (3)}$

Decimal Form:

$x = 0.38685280 \dots$