Find the Inverse f(x) = natural log of 5x-1

$f (x) = \ln (5 x - 1)$

Step 1

Write $f (x) = \ln (5 x - 1)$ as an equation.

$y = \ln (5 x - 1)$

Step 2

Interchange the variables.

$x = \ln (5 y - 1)$

Step 3

Solve for $y$ .

Rewrite the equation as $\ln (5 y - 1) = x$ .

$\ln (5 y - 1) = x$

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (5 y - 1)} = e^{x}$

Rewrite $\ln (5 y - 1) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{x} = 5 y - 1$

Solve for $y$ .

Rewrite the equation as $5 y - 1 = e^{x}$ .

$5 y - 1 = e^{x}$

Add $1$ to both sides of the equation.

$5 y = e^{x} + 1$

Divide each term in $5 y = e^{x} + 1$ by $5$ and simplify.

Divide each term in $5 y = e^{x} + 1$ by $5$ .

$\frac{5 y}{5} = \frac{e^{x}}{5} + \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y}{5} = \frac{e^{x}}{5} + \frac{1}{5}$

Divide $y$ by $1$ .

$y = \frac{e^{x}}{5} + \frac{1}{5}$

Step 4

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

$f^{- 1} (x) = \frac{e^{x}}{5} + \frac{1}{5}$

Step 5

Verify if $f^{- 1} (x) = \frac{e^{x}}{5} + \frac{1}{5}$ is the inverse of $f (x) = \ln (5 x - 1)$ .

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

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

Set up the composite result function.

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

Evaluate $f^{- 1} (\ln (5 x - 1))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\ln (5 x - 1)) = \frac{e^{\ln (5 x - 1)}}{5} + \frac{1}{5}$

Combine the numerators over the common denominator.

$f^{- 1} (\ln (5 x - 1)) = \frac{e^{\ln (5 x - 1)} + 1}{5}$

Exponentiation and log are inverse functions.

$f^{- 1} (\ln (5 x - 1)) = \frac{5 x - 1 + 1}{5}$

Combine the opposite terms in $5 x - 1 + 1$ .

Add $- 1$ and $1$ .

$f^{- 1} (\ln (5 x - 1)) = \frac{5 x + 0}{5}$

Add $5 x$ and $0$ .

$f^{- 1} (\ln (5 x - 1)) = \frac{5 x}{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\ln (5 x - 1)) = \frac{5 x}{5}$

Divide $x$ by $1$ .

$f^{- 1} (\ln (5 x - 1)) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{e^{x}}{5} + \frac{1}{5})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{e^{x}}{5} + \frac{1}{5}) = \ln (5 (\frac{e^{x}}{5} + \frac{1}{5}) - 1)$

Simplify each term.

Apply the distributive property.

$f (\frac{e^{x}}{5} + \frac{1}{5}) = \ln (5 (\frac{e^{x}}{5}) + 5 (\frac{1}{5}) - 1)$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\frac{e^{x}}{5} + \frac{1}{5}) = \ln (5 (\frac{e^{x}}{5}) + 5 (\frac{1}{5}) - 1)$

Rewrite the expression.

$f (\frac{e^{x}}{5} + \frac{1}{5}) = \ln (e^{x} + 5 (\frac{1}{5}) - 1)$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\frac{e^{x}}{5} + \frac{1}{5}) = \ln (e^{x} + 5 (\frac{1}{5}) - 1)$

Rewrite the expression.

$f (\frac{e^{x}}{5} + \frac{1}{5}) = \ln (e^{x} + 1 - 1)$

Combine the opposite terms in $e^{x} + 1 - 1$ .

Subtract $1$ from $1$ .

$f (\frac{e^{x}}{5} + \frac{1}{5}) = \ln (e^{x} + 0)$

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

$f (\frac{e^{x}}{5} + \frac{1}{5}) = \ln (e^{x})$

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

$f (\frac{e^{x}}{5} + \frac{1}{5}) = x \ln (e)$

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

$f (\frac{e^{x}}{5} + \frac{1}{5}) = x \cdot 1$

Multiply $x$ by $1$ .

$f (\frac{e^{x}}{5} + \frac{1}{5}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{e^{x}}{5} + \frac{1}{5}$ is the inverse of $f (x) = \ln (5 x - 1)$ .

$f^{- 1} (x) = \frac{e^{x}}{5} + \frac{1}{5}$

Find the Inverse f(x) = natural log of 5x-1

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