Trigonometry Examples

$f (x) = 2^{x + 3} - 1$

Step 1

Write $f (x) = 2^{x + 3} - 1$ as an equation.

$y = 2^{x + 3} - 1$

Step 2

Interchange the variables.

$x = 2^{y + 3} - 1$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $2^{y + 3} - 1 = x$ .

$2^{y + 3} - 1 = x$

Step 3.2

Add $1$ to both sides of the equation.

$2^{y + 3} = x + 1$

Step 3.3

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

$\ln (2^{y + 3}) = \ln (x + 1)$

Step 3.4

Expand $\ln (2^{y + 3})$ by moving $y + 3$ outside the logarithm.

$(y + 3) \ln (2) = \ln (x + 1)$

Step 3.5

Simplify the left side.

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Step 3.5.1

Apply the distributive property.

$y \ln (2) + 3 \ln (2) = \ln (x + 1)$

Step 3.6

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

$y \ln (2) + 3 \ln (2) - \ln (x + 1) = 0$

Step 3.7

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

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Step 3.7.1

Subtract $3 \ln (2)$ from both sides of the equation.

$y \ln (2) - \ln (x + 1) = - 3 \ln (2)$

Step 3.7.2

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

$y \ln (2) = - 3 \ln (2) + \ln (x + 1)$

Step 3.8

Divide each term in $y \ln (2) = - 3 \ln (2) + \ln (x + 1)$ by $\ln (2)$ and simplify.

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Step 3.8.1

Divide each term in $y \ln (2) = - 3 \ln (2) + \ln (x + 1)$ by $\ln (2)$ .

$\frac{y \ln (2)}{\ln (2)} = \frac{- 3 \ln (2)}{\ln (2)} + \frac{\ln (x + 1)}{\ln (2)}$

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.

$\frac{y \ln (2)}{\ln (2)} = \frac{- 3 \ln (2)}{\ln (2)} + \frac{\ln (x + 1)}{\ln (2)}$

Step 3.8.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 3 \ln (2)}{\ln (2)} + \frac{\ln (x + 1)}{\ln (2)}$

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.

$y = \frac{- 3 \ln (2)}{\ln (2)} + \frac{\ln (x + 1)}{\ln (2)}$

Step 3.8.3.1.2

Divide $- 3$ by $1$ .

$y = - 3 + \frac{\ln (x + 1)}{\ln (2)}$

Step 4

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

$f^{- 1} (x) = - 3 + \frac{\ln (x + 1)}{\ln (2)}$

Step 5

Verify if $f^{- 1} (x) = - 3 + \frac{\ln (x + 1)}{\ln (2)}$ is the inverse of $f (x) = 2^{x + 3} - 1$ .

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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 $f^{- 1} (2^{x + 3} - 1)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2^{x + 3} - 1) = - 3 + \frac{\ln ((2^{x + 3} - 1) + 1)}{\ln (2)}$

Step 5.2.3

Combine the opposite terms in $- 3 + \frac{\ln ((2^{x + 3} - 1) + 1)}{\ln (2)}$ .

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Step 5.2.3.1

Add $- 1$ and $1$ .

$f^{- 1} (2^{x + 3} - 1) = - 3 + \frac{\ln (2^{x + 3} + 0)}{\ln (2)}$

Step 5.2.3.2

Add $2^{x + 3}$ and $0$ .

$f^{- 1} (2^{x + 3} - 1) = - 3 + \frac{\ln (2^{x + 3})}{\ln (2)}$

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Expand $\ln (2^{x + 3})$ by moving $x + 3$ outside the logarithm.

$f^{- 1} (2^{x + 3} - 1) = - 3 + \frac{(x + 3) \ln (2)}{\ln (2)}$

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.

$f^{- 1} (2^{x + 3} - 1) = - 3 + \frac{(x + 3) \ln (2)}{\ln (2)}$

Step 5.2.4.2.2

Divide $x + 3$ by $1$ .

$f^{- 1} (2^{x + 3} - 1) = - 3 + x + 3$

Step 5.2.5

Combine the opposite terms in $- 3 + x + 3$ .

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Step 5.2.5.1

Add $- 3$ and $3$ .

$f^{- 1} (2^{x + 3} - 1) = x + 0$

Step 5.2.5.2

Add $x$ and $0$ .

$f^{- 1} (2^{x + 3} - 1) = x$

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 $f (- 3 + \frac{\ln (x + 1)}{\ln (2)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- 3 + \frac{\ln (x + 1)}{\ln (2)}) = 2^{(- 3 + \frac{\ln (x + 1)}{\ln (2)}) + 3} - 1$

Step 5.3.3

Combine the opposite terms in $2^{(- 3 + \frac{\ln (x + 1)}{\ln (2)}) + 3} - 1$ .

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Step 5.3.3.1

Add $- 3$ and $3$ .

$f (- 3 + \frac{\ln (x + 1)}{\ln (2)}) = 2^{0 + \frac{\ln (x + 1)}{\ln (2)}} - 1$

Step 5.3.3.2

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

$f (- 3 + \frac{\ln (x + 1)}{\ln (2)}) = 2^{\frac{\ln (x + 1)}{\ln (2)}} - 1$

Step 5.3.4

Simplify each term.

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Step 5.3.4.1

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

$f (- 3 + \frac{\ln (x + 1)}{\ln (2)}) = 2^{\log_{2} (x + 1)} - 1$

Step 5.3.4.2

Exponentiation and log are inverse functions.

$f (- 3 + \frac{\ln (x + 1)}{\ln (2)}) = x + 1 - 1$

Step 5.3.5

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

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Step 5.3.5.1

Subtract $1$ from $1$ .

$f (- 3 + \frac{\ln (x + 1)}{\ln (2)}) = x + 0$

Step 5.3.5.2

Add $x$ and $0$ .

$f (- 3 + \frac{\ln (x + 1)}{\ln (2)}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - 3 + \frac{\ln (x + 1)}{\ln (2)}$ is the inverse of $f (x) = 2^{x + 3} - 1$ .

$f^{- 1} (x) = - 3 + \frac{\ln (x + 1)}{\ln (2)}$