Find the Inverse -2 natural log of x+1

$- 2 \ln (x + 1)$

Interchange the variables.

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

Solve for $y$ .

Rewrite the equation as $- 2 \ln (y + 1) = x$ .

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

Divide each term in $- 2 \ln (y + 1) = x$ by $- 2$ and simplify.

Divide each term in $- 2 \ln (y + 1) = x$ by $- 2$ .

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $\ln (y + 1)$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (y + 1)} = e^{- \frac{x}{2}}$

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

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

Solve for $y$ .

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

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

Subtract $1$ from both sides of the equation.

$y = e^{- \frac{x}{2}} - 1$

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

$f^{- 1} (x) = e^{- \frac{x}{2}} - 1$

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

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

$f^{- 1} (- 2 \ln (x + 1)) = e^{- \frac{- 2 \ln (x + 1)}{2}} - 1$

Simplify each term.

Cancel the common factor of $- 2$ and $2$ .

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

$f^{- 1} (- 2 \ln (x + 1)) = e^{- \frac{2 (- \ln (x + 1))}{2}} - 1$

Cancel the common factors.

Factor $2$ out of $2$ .

$f^{- 1} (- 2 \ln (x + 1)) = e^{- \frac{2 (- \ln (x + 1))}{2 (1)}} - 1$

Cancel the common factor.

$f^{- 1} (- 2 \ln (x + 1)) = e^{- \frac{2 (- \ln (x + 1))}{2 \cdot 1}} - 1$

Rewrite the expression.

$f^{- 1} (- 2 \ln (x + 1)) = e^{- \frac{- \ln (x + 1)}{1}} - 1$

Divide $- \ln (x + 1)$ by $1$ .

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

Multiply $- - \ln (x + 1)$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $\ln (x + 1)$ by $1$ .

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

Exponentiation and log are inverse functions.

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

$f (e^{- \frac{x}{2}} - 1) = - 2 \ln ((e^{- \frac{x}{2}} - 1) + 1)$

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

Add $- 1$ and $1$ .

$f (e^{- \frac{x}{2}} - 1) = - 2 \ln (e^{- \frac{x}{2}} + 0)$

Add $e^{- \frac{x}{2}}$ and $0$ .

$f (e^{- \frac{x}{2}} - 1) = - 2 \ln (e^{- \frac{x}{2}})$

Use logarithm rules to move $- \frac{x}{2}$ out of the exponent.

$f (e^{- \frac{x}{2}} - 1) = - 2 (- \frac{x}{2} \cdot \ln (e))$

The natural logarithm of $e$ is $1$ .

$f (e^{- \frac{x}{2}} - 1) = - 2 (- \frac{x}{2} \cdot 1)$

Multiply $- 1$ by $1$ .

$f (e^{- \frac{x}{2}} - 1) = - 2 (- \frac{x}{2})$

Cancel the common factor of $2$ .

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

$f (e^{- \frac{x}{2}} - 1) = - 2 \frac{- x}{2}$

Factor $2$ out of $- 2$ .

$f (e^{- \frac{x}{2}} - 1) = 2 (- 1) (\frac{- x}{2})$

Cancel the common factor.

$f (e^{- \frac{x}{2}} - 1) = 2 \cdot (- 1 \frac{- x}{2})$

Rewrite the expression.

$f (e^{- \frac{x}{2}} - 1) = - 1 (- x)$

Multiply.

Multiply $- 1$ by $- 1$ .

$f (e^{- \frac{x}{2}} - 1) = 1 x$

Multiply $x$ by $1$ .

$f (e^{- \frac{x}{2}} - 1) = x$

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

$f^{- 1} (x) = e^{- \frac{x}{2}} - 1$

Find the Inverse -2 natural log of x+1

Related Questions

Support LCMGCF.com by disabling your adblocker.