Find the Inverse y=e^(2x+5)

$y = e^{2 x + 5}$

Step 1

Interchange the variables.

$x = e^{2 y + 5}$

Step 2

Solve for $y$ .

Rewrite the equation as $e^{2 y + 5} = x$ .

$e^{2 y + 5} = x$

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

$\ln (e^{2 y + 5}) = \ln (x)$

Expand the left side.

Expand $\ln (e^{2 y + 5})$ by moving $2 y + 5$ outside the logarithm.

$(2 y + 5) \ln (e) = \ln (x)$

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

$(2 y + 5) \cdot 1 = \ln (x)$

Multiply $2 y + 5$ by $1$ .

$2 y + 5 = \ln (x)$

Subtract $5$ from both sides of the equation.

$2 y = \ln (x) - 5$

Divide each term in $2 y = \ln (x) - 5$ by $2$ and simplify.

Divide each term in $2 y = \ln (x) - 5$ by $2$ .

$\frac{2 y}{2} = \frac{\ln (x)}{2} + \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{\ln (x)}{2} + \frac{- 5}{2}$

Divide $y$ by $1$ .

$y = \frac{\ln (x)}{2} + \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$y = \frac{\ln (x)}{2} - \frac{5}{2}$

Step 3

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

$f^{- 1} (x) = \frac{\ln (x)}{2} - \frac{5}{2}$

Step 4

Verify if $f^{- 1} (x) = \frac{\ln (x)}{2} - \frac{5}{2}$ is the inverse of $f (x) = e^{2 x + 5}$ .

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 $f^{- 1} (e^{2 x + 5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (e^{2 x + 5}) = \frac{\ln (e^{2 x + 5})}{2} - \frac{5}{2}$

Combine the numerators over the common denominator.

$f^{- 1} (e^{2 x + 5}) = \frac{\ln (e^{2 x + 5}) - 5}{2}$

Simplify each term.

Use logarithm rules to move $2 x + 5$ out of the exponent.

$f^{- 1} (e^{2 x + 5}) = \frac{(2 x + 5) \ln (e) - 5}{2}$

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

$f^{- 1} (e^{2 x + 5}) = \frac{(2 x + 5) \cdot 1 - 5}{2}$

Multiply $2 x + 5$ by $1$ .

$f^{- 1} (e^{2 x + 5}) = \frac{2 x + 5 - 5}{2}$

Simplify terms.

Combine the opposite terms in $2 x + 5 - 5$ .

Subtract $5$ from $5$ .

$f^{- 1} (e^{2 x + 5}) = \frac{2 x + 0}{2}$

Add $2 x$ and $0$ .

$f^{- 1} (e^{2 x + 5}) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (e^{2 x + 5}) = \frac{2 x}{2}$

Divide $x$ by $1$ .

$f^{- 1} (e^{2 x + 5}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\ln (x)}{2} - \frac{5}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{2 (\frac{\ln (x)}{2} - \frac{5}{2}) + 5}$

Simplify each term.

Rewrite $\frac{\ln (x)}{2}$ as $\frac{1}{2} \ln (x)$ .

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{2 (\frac{1}{2} \cdot \ln (x) - \frac{5}{2}) + 5}$

Simplify $\frac{1}{2} \ln (x)$ by moving $\frac{1}{2}$ inside the logarithm.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{2 (\ln (x^{\frac{1}{2}}) - \frac{5}{2}) + 5}$

Apply the distributive property.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{2 \ln (x^{\frac{1}{2}}) + 2 (- \frac{5}{2}) + 5}$

Simplify $2 \ln (x^{\frac{1}{2}})$ by moving $2$ inside the logarithm.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln ({(x^{\frac{1}{2}})}^{2}) + 2 (- \frac{5}{2}) + 5}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{5}{2}$ into the numerator.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln ({(x^{\frac{1}{2}})}^{2}) + 2 (\frac{- 5}{2}) + 5}$

Cancel the common factor.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln ({(x^{\frac{1}{2}})}^{2}) + 2 (\frac{- 5}{2}) + 5}$

Rewrite the expression.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln ({(x^{\frac{1}{2}})}^{2}) - 5 + 5}$

Simplify each term.

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln (x^{\frac{1}{2} \cdot 2}) - 5 + 5}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln (x^{\frac{1}{2} \cdot 2}) - 5 + 5}$

Rewrite the expression.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln (x^{1}) - 5 + 5}$

Simplify.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln (x) - 5 + 5}$

Combine the opposite terms in $\ln (x) - 5 + 5$ .

Add $- 5$ and $5$ .

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln (x) + 0}$

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

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = e^{\ln (x)}$

Exponentiation and log are inverse functions.

$f (\frac{\ln (x)}{2} - \frac{5}{2}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x)}{2} - \frac{5}{2}$ is the inverse of $f (x) = e^{2 x + 5}$ .

$f^{- 1} (x) = \frac{\ln (x)}{2} - \frac{5}{2}$

Find the Inverse y=e^(2x+5)

Related Questions

Support LCMGCF.com by disabling your adblocker.