Trigonometry Examples

$y = e^{x - 1}$

Step 1

Interchange the variables.

$x = e^{y - 1}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $e^{y - 1} = x$ .

$e^{y - 1} = x$

Step 2.2

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

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

Step 2.3

Expand the left side.

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Step 2.3.1

Expand $\ln (e^{y - 1})$ by moving $y - 1$ outside the logarithm.

$(y - 1) \ln (e) = \ln (x)$

Step 2.3.2

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

$(y - 1) \cdot 1 = \ln (x)$

Step 2.3.3

Multiply $y - 1$ by $1$ .

$y - 1 = \ln (x)$

Step 2.4

Add $1$ to both sides of the equation.

$y = \ln (x) + 1$

Step 3

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

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

Step 4

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

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

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

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

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Use logarithm rules to move $x - 1$ out of the exponent.

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

Step 4.2.3.2

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

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

Step 4.2.3.3

Multiply $x - 1$ by $1$ .

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

Step 4.2.4

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

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Step 4.2.4.1

Add $- 1$ and $1$ .

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

Step 4.2.4.2

Add $x$ and $0$ .

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

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

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

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

Step 4.3.3

Combine the opposite terms in $(\ln (x) + 1) - 1$ .

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Step 4.3.3.1

Subtract $1$ from $1$ .

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

Step 4.3.3.2

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

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

Step 4.3.4

Exponentiation and log are inverse functions.

$f (\ln (x) + 1) = x$

Step 4.4

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

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