Algebra Examples

$y = \frac{1}{5} \cdot e^{x + 2}$

Step 1

Interchange the variables.

$x = \frac{1}{5} \cdot e^{y + 2}$

Step 2

Solve for $y$ .

Tap for more steps...

Step 2.1

Rewrite the equation as $\frac{1}{5} \cdot e^{y + 2} = x$ .

$\frac{1}{5} \cdot e^{y + 2} = x$

Step 2.2

Multiply both sides of the equation by $5$ .

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

Step 2.3

Simplify the left side.

Tap for more steps...

Step 2.3.1

Simplify $5 (\frac{1}{5} \cdot e^{y + 2})$ .

Tap for more steps...

Step 2.3.1.1

Combine $\frac{1}{5}$ and $e^{y + 2}$ .

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

Step 2.3.1.2

Cancel the common factor of $5$ .

Tap for more steps...

Step 2.3.1.2.1

Cancel the common factor.

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

Step 2.3.1.2.2

Rewrite the expression.

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

Step 2.4

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

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

Step 2.5

Expand the left side.

Tap for more steps...

Step 2.5.1

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

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

Step 2.5.2

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

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

Step 2.5.3

Multiply $y + 2$ by $1$ .

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

Step 2.6

Subtract $2$ from both sides of the equation.

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

Step 3

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

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

Step 4

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

Tap for more steps...

Step 4.1

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

Step 4.2

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

Tap for more steps...

Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

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

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

Step 4.2.3

Simplify each term.

Tap for more steps...

Step 4.2.3.1

Combine $\frac{1}{5}$ and $e^{x + 2}$ .

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

Step 4.2.3.2

Cancel the common factor of $5$ .

Tap for more steps...

Step 4.2.3.2.1

Cancel the common factor.

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

Step 4.2.3.2.2

Rewrite the expression.

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

Step 4.2.3.3

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

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

Step 4.2.3.4

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

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

Step 4.2.3.5

Multiply $x + 2$ by $1$ .

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

Step 4.2.4

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

Tap for more steps...

Step 4.2.4.1

Subtract $2$ from $2$ .

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

Step 4.2.4.2

Add $x$ and $0$ .

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

Step 4.3

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

Tap for more steps...

Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

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

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

Step 4.3.3

Combine the opposite terms in $(\ln (5 x) - 2) + 2$ .

Tap for more steps...

Step 4.3.3.1

Add $- 2$ and $2$ .

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

Step 4.3.3.2

Add $\ln (5 x)$ and $0$ .

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

Step 4.3.4

Exponentiation and log are inverse functions.

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

Step 4.3.5

Cancel the common factor of $5$ .

Tap for more steps...

Step 4.3.5.1

Factor $5$ out of $5 x$ .

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

Step 4.3.5.2

Cancel the common factor.

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

Step 4.3.5.3

Rewrite the expression.

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

Step 4.4

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

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