Find the Inverse 2- log base 3 of x+1

$2 - \log_{3} (x + 1)$

Interchange the variables.

$x = 2 - \log_{3} (y + 1)$

Solve for $y$ .

Rewrite the equation as $2 - \log_{3} (y + 1) = x$ .

$2 - \log_{3} (y + 1) = x$

Subtract $2$ from both sides of the equation.

$- \log_{3} (y + 1) = x - 2$

Divide each term in $- \log_{3} (y + 1) = x - 2$ by $- 1$ and simplify.

Divide each term in $- \log_{3} (y + 1) = x - 2$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $\log_{3} (y + 1)$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x}{- 1}$ .

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

Rewrite $- 1 \cdot x$ as $- x$ .

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

Divide $- 2$ by $- 1$ .

$\log_{3} (y + 1) = - x + 2$

Rewrite $\log_{3} (y + 1) = - x + 2$ 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 $b^{y} = x$ .

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

Solve for $y$ .

Rewrite the equation as $y + 1 = 3^{- x + 2}$ .

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

Subtract $1$ from both sides of the equation.

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

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

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

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

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 1$ by $2$ .

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

Multiply $- - \log_{3} (x + 1)$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $\log_{3} (x + 1)$ by $1$ .

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

Combine the opposite terms in $- 2 + \log_{3} (x + 1) + 2$ .

Add $- 2$ and $2$ .

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

Add $0$ and $\log_{3} (x + 1)$ .

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

Exponentiation and log are inverse functions.

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

Combine the opposite terms in $x + 1 - 1$ .

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Combine the opposite terms in $2 - \log_{3} ((3^{- x + 2} - 1) + 1)$ .

Add $- 1$ and $1$ .

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

Add $3^{- x + 2}$ and $0$ .

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

Simplify each term.

Use logarithm rules to move $- x + 2$ out of the exponent.

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

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

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

Multiply $- x + 2$ by $1$ .

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

Apply the distributive property.

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

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $2$ .

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

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

Subtract $2$ from $2$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse 2- log base 3 of x+1

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