Trigonometry Examples

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

Step 1

Write $f (x) = 2^{x - 1} + 1$ as an equation.

$y = 2^{x - 1} + 1$

Step 2

Interchange the variables.

$x = 2^{y - 1} + 1$

Step 3

Solve for $y$ .

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Step 3.1

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

$2^{y - 1} + 1 = x$

Step 3.2

Subtract $1$ from both sides of the equation.

$2^{y - 1} = x - 1$

Step 3.3

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

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

Step 3.4

Expand $\ln (2^{y - 1})$ by moving $y - 1$ outside the logarithm.

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

Step 3.5

Simplify the left side.

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Step 3.5.1

Simplify $(y - 1) \ln (2)$ .

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Step 3.5.1.1

Apply the distributive property.

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

Step 3.5.1.2

Rewrite $- 1 \ln (2)$ as $- \ln (2)$ .

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

Step 3.6

Move all the terms containing a logarithm to the left side of the equation.

$y \ln (2) - \ln (2) - \ln (x - 1) = 0$

Step 3.7

Move all terms not containing $y$ to the right side of the equation.

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Step 3.7.1

Add $\ln (2)$ to both sides of the equation.

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

Step 3.7.2

Add $\ln (x - 1)$ to both sides of the equation.

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

Step 3.8

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

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Step 3.8.1

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

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

Step 3.8.2

Simplify the left side.

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Step 3.8.2.1

Cancel the common factor of $\ln (2)$ .

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Step 3.8.2.1.1

Cancel the common factor.

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

Step 3.8.2.1.2

Divide $y$ by $1$ .

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

Step 3.8.3

Simplify the right side.

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Step 3.8.3.1

Cancel the common factor of $\ln (2)$ .

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Step 3.8.3.1.1

Cancel the common factor.

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

Step 3.8.3.1.2

Rewrite the expression.

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

Step 4

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

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

Step 5

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

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

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

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

Step 5.2.3

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

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Step 5.2.3.1

Subtract $1$ from $1$ .

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

Step 5.2.3.2

Add $2^{x - 1}$ and $0$ .

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

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Expand $\ln (2^{x - 1})$ by moving $x - 1$ outside the logarithm.

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

Step 5.2.4.2

Cancel the common factor of $\ln (2)$ .

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Step 5.2.4.2.1

Cancel the common factor.

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

Step 5.2.4.2.2

Divide $x - 1$ by $1$ .

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

Step 5.2.5

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

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Step 5.2.5.1

Subtract $1$ from $1$ .

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

Step 5.2.5.2

Add $x$ and $0$ .

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

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

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

Step 5.3.3

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

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Step 5.3.3.1

Subtract $1$ from $1$ .

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

Step 5.3.3.2

Add $0$ and $\frac{\ln (x - 1)}{\ln (2)}$ .

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

Step 5.3.4

Simplify each term.

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Step 5.3.4.1

Use the change of base rule $\frac{\log_{b} (x)}{\log_{b} (a)} = \log_{a} (x)$ .

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

Step 5.3.4.2

Exponentiation and log are inverse functions.

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

Step 5.3.5

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

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Step 5.3.5.1

Add $- 1$ and $1$ .

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

Step 5.3.5.2

Add $x$ and $0$ .

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

Step 5.4

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

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