Find the Inverse g(x) = log base 3 of 3-2x

$g (x) = \log_{3} (3 - 2 x)$

Step 1

Write $g (x) = \log_{3} (3 - 2 x)$ as an equation.

$y = \log_{3} (3 - 2 x)$

Step 2

Interchange the variables.

$x = \log_{3} (3 - 2 y)$

Step 3

Solve for $y$ .

Rewrite the equation as $\log_{3} (3 - 2 y) = x$ .

$\log_{3} (3 - 2 y) = x$

Rewrite $\log_{3} (3 - 2 y) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

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

Solve for $y$ .

Rewrite the equation as $3 - 2 y = 3^{x}$ .

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

Subtract $3$ from both sides of the equation.

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

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

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

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Dividing two negative values results in a positive value.

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

Step 4

Replace $y$ with $g^{- 1} (x)$ to show the final answer.

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

Step 5

Verify if $g^{- 1} (x) = - \frac{3^{x}}{2} + \frac{3}{2}$ is the inverse of $g (x) = \log_{3} (3 - 2 x)$ .

To verify the inverse, check if $g^{- 1} (g (x)) = x$ and $g (g^{- 1} (x)) = x$ .

Evaluate $g^{- 1} (g (x))$ .

Set up the composite result function.

$g^{- 1} (g (x))$

Evaluate $g^{- 1} (\log_{3} (3 - 2 x))$ by substituting in the value of $g$ into $g^{- 1}$ .

$g^{- 1} (\log_{3} (3 - 2 x)) = - \frac{3^{\log_{3} (3 - 2 x)}}{2} + \frac{3}{2}$

Combine the numerators over the common denominator.

$g^{- 1} (\log_{3} (3 - 2 x)) = \frac{- 3^{\log_{3} (3 - 2 x)} + 3}{2}$

Simplify each term.

Exponentiation and log are inverse functions.

$g^{- 1} (\log_{3} (3 - 2 x)) = \frac{- (3 - 2 x) + 3}{2}$

Apply the distributive property.

$g^{- 1} (\log_{3} (3 - 2 x)) = \frac{- 1 \cdot 3 - (- 2 x) + 3}{2}$

Multiply $- 1$ by $3$ .

$g^{- 1} (\log_{3} (3 - 2 x)) = \frac{- 3 - (- 2 x) + 3}{2}$

Multiply $- 2$ by $- 1$ .

$g^{- 1} (\log_{3} (3 - 2 x)) = \frac{- 3 + 2 x + 3}{2}$

Simplify terms.

Combine the opposite terms in $- 3 + 2 x + 3$ .

Add $- 3$ and $3$ .

$g^{- 1} (\log_{3} (3 - 2 x)) = \frac{2 x + 0}{2}$

Add $2 x$ and $0$ .

$g^{- 1} (\log_{3} (3 - 2 x)) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$g^{- 1} (\log_{3} (3 - 2 x)) = \frac{2 x}{2}$

Divide $x$ by $1$ .

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

Evaluate $g (g^{- 1} (x))$ .

Set up the composite result function.

$g (g^{- 1} (x))$

Evaluate $g (- \frac{3^{x}}{2} + \frac{3}{2})$ by substituting in the value of $g^{- 1}$ into $g$ .

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 - 2 (- \frac{3^{x}}{2} + \frac{3}{2}))$

Simplify each term.

Apply the distributive property.

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 - 2 (- \frac{3^{x}}{2}) - 2 (\frac{3}{2}))$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3^{x}}{2}$ into the numerator.

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 - 2 \frac{- 3^{x}}{2} - 2 (\frac{3}{2}))$

Factor $2$ out of $- 2$ .

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 + 2 (- 1) (\frac{- 3^{x}}{2}) - 2 (\frac{3}{2}))$

Cancel the common factor.

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 + 2 \cdot (- 1 \frac{- 3^{x}}{2}) - 2 (\frac{3}{2}))$

Rewrite the expression.

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 - 1 (- 3^{x}) - 2 (\frac{3}{2}))$

Multiply $- 1$ by $- 1$ .

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 + 1 \cdot 3^{x} - 2 (\frac{3}{2}))$

Multiply $3^{x}$ by $1$ .

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 + 3^{x} - 2 (\frac{3}{2}))$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 + 3^{x} + 2 (- 1) (\frac{3}{2}))$

Cancel the common factor.

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 + 3^{x} + 2 \cdot (- 1 (\frac{3}{2})))$

Rewrite the expression.

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 + 3^{x} - 1 \cdot 3)$

Multiply $- 1$ by $3$ .

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3 + 3^{x} - 3)$

Combine the opposite terms in $3 + 3^{x} - 3$ .

Subtract $3$ from $3$ .

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (0 + 3^{x})$

Add $0$ and $3^{x}$ .

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = \log_{3} (3^{x})$

Use logarithm rules to move $x$ out of the exponent.

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = x \log_{3} (3)$

Logarithm base $3$ of $3$ is $1$ .

$g (- \frac{3^{x}}{2} + \frac{3}{2}) = x \cdot 1$

Multiply $x$ by $1$ .

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

Since $g^{- 1} (g (x)) = x$ and $g (g^{- 1} (x)) = x$ , then $g^{- 1} (x) = - \frac{3^{x}}{2} + \frac{3}{2}$ is the inverse of $g (x) = \log_{3} (3 - 2 x)$ .

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

Find the Inverse g(x) = log base 3 of 3-2x

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