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

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

Step 1

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

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

Step 2

Solve for $x$ .

Subtract $2$ from both sides of the equation.

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

Divide each term in $- 3^{2 x + 1} = - 2$ by $- 1$ and simplify.

Divide each term in $- 3^{2 x + 1} = - 2$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $3^{2 x + 1}$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $- 1$ .

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

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Expand $\ln (3^{2 x + 1})$ by moving $2 x + 1$ outside the logarithm.

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

Simplify the left side.

Simplify $(2 x + 1) \ln (3)$ .

Apply the distributive property.

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

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

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

Move all the terms containing a logarithm to the left side of the equation.

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

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$2 x \ln (3) + \ln (\frac{3}{2}) = 0$

Subtract $\ln (\frac{3}{2})$ from both sides of the equation.

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

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

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

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