Trigonometry Examples

$f (x) = {2.3}^{x}$

Step 1

Write $f (x) = {2.3}^{x}$ as an equation.

$y = {2.3}^{x}$

Step 2

Interchange the variables.

$x = {2.3}^{y}$

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

Rewrite the equation as ${2.3}^{y} = x$ .

${2.3}^{y} = x$

Step 3.2

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

$\ln ({2.3}^{y}) = \ln (x)$

Step 3.3

Expand $\ln ({2.3}^{y})$ by moving $y$ outside the logarithm.

$y \ln (2.3) = \ln (x)$

Step 3.4

Divide each term in $y \ln (2.3) = \ln (x)$ by $\ln (2.3)$ and simplify.

Tap for more steps...

Step 3.4.1

Divide each term in $y \ln (2.3) = \ln (x)$ by $\ln (2.3)$ .

$\frac{y \ln (2.3)}{\ln (2.3)} = \frac{\ln (x)}{\ln (2.3)}$

Step 3.4.2

Simplify the left side.

Tap for more steps...

Step 3.4.2.1

Cancel the common factor of $\ln (2.3)$ .

Tap for more steps...

Step 3.4.2.1.1

Cancel the common factor.

$\frac{y \ln (2.3)}{\ln (2.3)} = \frac{\ln (x)}{\ln (2.3)}$

Step 3.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (x)}{\ln (2.3)}$

Step 4

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

$f^{- 1} (x) = \frac{\ln (x)}{\ln (2.3)}$

Step 5

Verify if $f^{- 1} (x) = \frac{\ln (x)}{\ln (2.3)}$ is the inverse of $f (x) = {2.3}^{x}$ .

Tap for more steps...

Step 5.1

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

Step 5.2

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

Tap for more steps...

Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate $f^{- 1} \cdot {2.3}^{x}$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} \cdot {2.3}^{x} = \frac{\ln ({2.3}^{x})}{\ln (2.3)}$

Step 5.2.3

Expand $\ln ({2.3}^{x})$ by moving $x$ outside the logarithm.

$f^{- 1} \cdot {2.3}^{x} = \frac{x \ln (2.3)}{\ln (2.3)}$

Step 5.2.4

Cancel the common factor of $\ln (2.3)$ .

Tap for more steps...

Step 5.2.4.1

Cancel the common factor.

$f^{- 1} \cdot {2.3}^{x} = \frac{x \ln (2.3)}{\ln (2.3)}$

Step 5.2.4.2

Divide $x$ by $1$ .

$f^{- 1} \cdot {2.3}^{x} = x$

Step 5.3

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

Tap for more steps...

Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate $f (\frac{\ln (x)}{\ln (2.3)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (x)}{\ln (2.3)}) = {2.3}^{\frac{\ln (x)}{\ln (2.3)}}$

Step 5.3.3

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

$f (\frac{\ln (x)}{\ln (2.3)}) = {2.3}^{\log_{2.3} (x)}$

Step 5.3.4

Exponentiation and log are inverse functions.

$f (\frac{\ln (x)}{\ln (2.3)}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (x)}{\ln (2.3)}$ is the inverse of $f (x) = {2.3}^{x}$ .

$f^{- 1} (x) = \frac{\ln (x)}{\ln (2.3)}$