Algebra Examples

$f (x) = {1.5}^{x} + 4$

Step 1

Write $f (x) = {1.5}^{x} + 4$ as an equation.

$y = {1.5}^{x} + 4$

Step 2

Interchange the variables.

$x = {1.5}^{y} + 4$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as ${1.5}^{y} + 4 = x$ .

${1.5}^{y} + 4 = x$

Step 3.2

Subtract $4$ from both sides of the equation.

${1.5}^{y} = x - 4$

Step 3.3

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

$\ln ({1.5}^{y}) = \ln (x - 4)$

Step 3.4

Expand $\ln ({1.5}^{y})$ by moving $y$ outside the logarithm.

$y \ln (1.5) = \ln (x - 4)$

Step 3.5

Divide each term in $y \ln (1.5) = \ln (x - 4)$ by $\ln (1.5)$ and simplify.

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Step 3.5.1

Divide each term in $y \ln (1.5) = \ln (x - 4)$ by $\ln (1.5)$ .

$\frac{y \ln (1.5)}{\ln (1.5)} = \frac{\ln (x - 4)}{\ln (1.5)}$

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Cancel the common factor of $\ln (1.5)$ .

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Step 3.5.2.1.1

Cancel the common factor.

$\frac{y \ln (1.5)}{\ln (1.5)} = \frac{\ln (x - 4)}{\ln (1.5)}$

Step 3.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (x - 4)}{\ln (1.5)}$

Step 4

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

$f^{- 1} (x) = \frac{\ln (x - 4)}{\ln (1.5)}$

Step 5

Verify if $f^{- 1} (x) = \frac{\ln (x - 4)}{\ln (1.5)}$ is the inverse of $f (x) = {1.5}^{x} + 4$ .

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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 $f^{- 1} ({1.5}^{x} + 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ({1.5}^{x} + 4) = \frac{\ln (({1.5}^{x} + 4) - 4)}{\ln (1.5)}$

Step 5.2.3

Combine the opposite terms in ${1.5}^{x} + 4 - 4$ .

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Step 5.2.3.1

Subtract $4$ from $4$ .

$f^{- 1} ({1.5}^{x} + 4) = \frac{\ln ({1.5}^{x} + 0)}{\ln (1.5)}$

Step 5.2.3.2

Add ${1.5}^{x}$ and $0$ .

$f^{- 1} ({1.5}^{x} + 4) = \frac{\ln ({1.5}^{x})}{\ln (1.5)}$

Step 5.2.4

Expand $\ln ({1.5}^{x})$ by moving $x$ outside the logarithm.

$f^{- 1} ({1.5}^{x} + 4) = \frac{x \ln (1.5)}{\ln (1.5)}$

Step 5.2.5

Cancel the common factor of $\ln (1.5)$ .

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Step 5.2.5.1

Cancel the common factor.

$f^{- 1} ({1.5}^{x} + 4) = \frac{x \ln (1.5)}{\ln (1.5)}$

Step 5.2.5.2

Divide $x$ by $1$ .

$f^{- 1} ({1.5}^{x} + 4) = x$

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 $f (\frac{\ln (x - 4)}{\ln (1.5)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x - 4)}{\ln (1.5)}) = {1.5}^{\frac{\ln (x - 4)}{\ln (1.5)}} + 4$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

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

$f (\frac{\ln (x - 4)}{\ln (1.5)}) = {1.5}^{\log_{1.5} (x - 4)} + 4$

Step 5.3.3.2

Exponentiation and log are inverse functions.

$f (\frac{\ln (x - 4)}{\ln (1.5)}) = x - 4 + 4$

Step 5.3.4

Combine the opposite terms in $x - 4 + 4$ .

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Step 5.3.4.1

Add $- 4$ and $4$ .

$f (\frac{\ln (x - 4)}{\ln (1.5)}) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (\frac{\ln (x - 4)}{\ln (1.5)}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x - 4)}{\ln (1.5)}$ is the inverse of $f (x) = {1.5}^{x} + 4$ .

$f^{- 1} (x) = \frac{\ln (x - 4)}{\ln (1.5)}$