Find the Roots (Zeros) f(x)=2^(x-1)-3^(x+1)

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

Step 1

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

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

Step 2

Solve for $x$ .

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

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

Take the log of both sides of the equation.

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

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

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

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

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

Solve the equation for $x$ .

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

Start simplifying.

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

Simplify by adding zeros.

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

Apply the distributive property.

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

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

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

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

Apply the distributive property.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

Step 3

Find the Roots (Zeros) f(x)=2^(x-1)-3^(x+1)

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