Algebra Examples

$f (x) = 2^{x - 1} - 3^{x + 1}$

Step 1

Set $2^{x - 1} - 3^{x + 1}$ equal to $0$ .

$2^{x - 1} - 3^{x + 1} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Move $- 3^{x + 1}$ to the right side of the equation by adding it to both sides.

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

Step 2.2

Take the log of both sides of the equation.

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Solve the equation for $x$ .

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Step 2.5.1

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

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Step 2.5.1.1

Rewrite.

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

Step 2.5.1.2

Simplify by adding zeros.

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

Step 2.5.1.3

Apply the distributive property.

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

Step 2.5.1.4

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

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

Step 2.5.2

Simplify $(x + 1) \ln (3)$ .

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Step 2.5.2.1

Apply the distributive property.

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

Step 2.5.2.2

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

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

Step 2.5.3

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

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

Step 2.5.4

Add $\ln (2)$ to both sides of the equation.

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

Step 2.5.5

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

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Step 2.5.5.1

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

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

Step 2.5.5.2

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

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

Step 2.5.5.3

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

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

Step 2.5.6

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

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

Step 2.5.7

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

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Step 2.5.7.1

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

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

Step 2.5.7.2

Simplify the left side.

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Step 2.5.7.2.1

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

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Step 2.5.7.2.1.1

Cancel the common factor.

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

Step 2.5.7.2.1.2

Divide $x$ by $1$ .

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

Step 2.5.7.3

Simplify the right side.

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Step 2.5.7.3.1

Combine the numerators over the common denominator.

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

Step 3