Find the Inverse natural log of 2x+4

$\ln (2 x) + 4$

Interchange the variables.

$x = \ln (2 y) + 4$

Solve for $y$ .

Rewrite the equation as $\ln (2 y) + 4 = x$ .

$\ln (2 y) + 4 = x$

Subtract $4$ from both sides of the equation.

$\ln (2 y) = x - 4$

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

$e^{\ln (2 y)} = e^{x - 4}$

Rewrite $\ln (2 y) = x - 4$ 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 $b^{y} = x$ .

$e^{x - 4} = 2 y$

Solve for $y$ .

Rewrite the equation as $2 y = e^{x - 4}$ .

$2 y = e^{x - 4}$

Divide each term in $2 y = e^{x - 4}$ by $2$ and simplify.

Divide each term in $2 y = e^{x - 4}$ by $2$ .

$\frac{2 y}{2} = \frac{e^{x - 4}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{e^{x - 4}}{2}$

Divide $y$ by $1$ .

$y = \frac{e^{x - 4}}{2}$

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

$f^{- 1} (x) = \frac{e^{x - 4}}{2}$

Verify if $f^{- 1} (x) = \frac{e^{x - 4}}{2}$ is the inverse of $f (x) = \ln (2 x) + 4$ .

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 $f^{- 1} (\ln (2 x) + 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\ln (2 x) + 4) = \frac{e^{(\ln (2 x) + 4) - 4}}{2}$

Simplify the numerator.

Combine the opposite terms in $\ln (2 x) + 4 - 4$ .

Subtract $4$ from $4$ .

$f^{- 1} (\ln (2 x) + 4) = \frac{e^{\ln (2 x) + 0}}{2}$

Add $\ln (2 x)$ and $0$ .

$f^{- 1} (\ln (2 x) + 4) = \frac{e^{\ln (2 x)}}{2}$

Exponentiation and log are inverse functions.

$f^{- 1} (\ln (2 x) + 4) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\ln (2 x) + 4) = \frac{2 x}{2}$

Divide $x$ by $1$ .

$f^{- 1} (\ln (2 x) + 4) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{e^{x - 4}}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{e^{x - 4}}{2}) = \ln (2 (\frac{e^{x - 4}}{2})) + 4$

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\frac{e^{x - 4}}{2}) = \ln (2 (\frac{e^{x - 4}}{2})) + 4$

Rewrite the expression.

$f (\frac{e^{x - 4}}{2}) = \ln (e^{x - 4}) + 4$

Use logarithm rules to move $x - 4$ out of the exponent.

$f (\frac{e^{x - 4}}{2}) = (x - 4) \ln (e) + 4$

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

$f (\frac{e^{x - 4}}{2}) = (x - 4) \cdot 1 + 4$

Multiply $x - 4$ by $1$ .

$f (\frac{e^{x - 4}}{2}) = x - 4 + 4$

Combine the opposite terms in $x - 4 + 4$ .

Add $- 4$ and $4$ .

$f (\frac{e^{x - 4}}{2}) = x + 0$

Add $x$ and $0$ .

$f (\frac{e^{x - 4}}{2}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{e^{x - 4}}{2}$ is the inverse of $f (x) = \ln (2 x) + 4$ .

$f^{- 1} (x) = \frac{e^{x - 4}}{2}$

Find the Inverse natural log of 2x+4

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