Find the Inverse y=3e^(-4x+1)

$y = 3 e^{- 4 x + 1}$

Step 1

Interchange the variables.

$x = 3 e^{- 4 y + 1}$

Step 2

Solve for $y$ .

Rewrite the equation as $3 e^{- 4 y + 1} = x$ .

$3 e^{- 4 y + 1} = x$

Divide each term in $3 e^{- 4 y + 1} = x$ by $3$ and simplify.

Divide each term in $3 e^{- 4 y + 1} = x$ by $3$ .

$\frac{3 e^{- 4 y + 1}}{3} = \frac{x}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 e^{- 4 y + 1}}{3} = \frac{x}{3}$

Divide $e^{- 4 y + 1}$ by $1$ .

$e^{- 4 y + 1} = \frac{x}{3}$

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

$\ln (e^{- 4 y + 1}) = \ln (\frac{x}{3})$

Expand the left side.

Expand $\ln (e^{- 4 y + 1})$ by moving $- 4 y + 1$ outside the logarithm.

$(- 4 y + 1) \ln (e) = \ln (\frac{x}{3})$

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

$(- 4 y + 1) \cdot 1 = \ln (\frac{x}{3})$

Multiply $- 4 y + 1$ by $1$ .

$- 4 y + 1 = \ln (\frac{x}{3})$

Subtract $1$ from both sides of the equation.

$- 4 y = \ln (\frac{x}{3}) - 1$

Divide each term in $- 4 y = \ln (\frac{x}{3}) - 1$ by $- 4$ and simplify.

Divide each term in $- 4 y = \ln (\frac{x}{3}) - 1$ by $- 4$ .

$\frac{- 4 y}{- 4} = \frac{\ln (\frac{x}{3})}{- 4} + \frac{- 1}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 y}{- 4} = \frac{\ln (\frac{x}{3})}{- 4} + \frac{- 1}{- 4}$

Divide $y$ by $1$ .

$y = \frac{\ln (\frac{x}{3})}{- 4} + \frac{- 1}{- 4}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{\ln (\frac{x}{3})}{4} + \frac{- 1}{- 4}$

Dividing two negative values results in a positive value.

$y = - \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}$

Step 3

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

$f^{- 1} (x) = - \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}$

Step 4

Verify if $f^{- 1} (x) = - \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}$ is the inverse of $f (x) = 3 e^{- 4 x + 1}$ .

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 $f^{- 1} (3 e^{- 4 x + 1})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (3 e^{- 4 x + 1}) = - \frac{\ln (\frac{3 e^{- 4 x + 1}}{3})}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{- \ln (\frac{3 e^{- 4 x + 1}}{3}) + 1}{4}$

Simplify each term.

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{- \ln (\frac{3 e^{- 4 x + 1}}{3}) + 1}{4}$

Divide $e^{- 4 x + 1}$ by $1$ .

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{- \ln (e^{- 4 x + 1}) + 1}{4}$

Use logarithm rules to move $- 4 x + 1$ out of the exponent.

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{- ((- 4 x + 1) \ln (e)) + 1}{4}$

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

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{- ((- 4 x + 1) \cdot 1) + 1}{4}$

Multiply $- 4 x + 1$ by $1$ .

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{- (- 4 x + 1) + 1}{4}$

Apply the distributive property.

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{- (- 4 x) - 1 \cdot 1 + 1}{4}$

Multiply $- 4$ by $- 1$ .

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{4 x - 1 \cdot 1 + 1}{4}$

Multiply $- 1$ by $1$ .

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{4 x - 1 + 1}{4}$

Simplify terms.

Combine the opposite terms in $4 x - 1 + 1$ .

Add $- 1$ and $1$ .

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{4 x + 0}{4}$

Add $4 x$ and $0$ .

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{4 x}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (3 e^{- 4 x + 1}) = \frac{4 x}{4}$

Divide $x$ by $1$ .

$f^{- 1} (3 e^{- 4 x + 1}) = x$

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

Set up the composite result function.

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

Evaluate $f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{- 4 (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) + 1}$

Simplify each term.

Rewrite $\frac{\ln (\frac{x}{3})}{4}$ as $\frac{1}{4} \ln (\frac{1}{3} x)$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{- 4 (- (\frac{1}{4} \cdot \ln (\frac{1}{3} x)) + \frac{1}{4}) + 1}$

Simplify $\frac{1}{4} \ln (\frac{1}{3} x)$ by moving $\frac{1}{4}$ inside the logarithm.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{- 4 (- \ln ({(\frac{1}{3} x)}^{\frac{1}{4}}) + \frac{1}{4}) + 1}$

Combine $\frac{1}{3}$ and $x$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{- 4 (- \ln ({(\frac{x}{3})}^{\frac{1}{4}}) + \frac{1}{4}) + 1}$

Apply the product rule to $\frac{x}{3}$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{- 4 (- \ln (\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}}) + \frac{1}{4}) + 1}$

Apply the distributive property.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{- 4 (- \ln (\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}})) - 4 (\frac{1}{4}) + 1}$

Multiply $- 4 (- \ln (\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}}))$ .

Multiply $- 1$ by $- 4$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{4 \ln (\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}}) - 4 (\frac{1}{4}) + 1}$

Simplify $4 \ln (\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}})$ by moving $4$ inside the logarithm.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln ({(\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}})}^{4}) - 4 (\frac{1}{4}) + 1}$

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln ({(\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}})}^{4}) + 4 (- 1) (\frac{1}{4}) + 1}$

Cancel the common factor.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln ({(\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}})}^{4}) + 4 \cdot (- 1 (\frac{1}{4})) + 1}$

Rewrite the expression.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln ({(\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}})}^{4}) - 1 + 1}$

Simplify each term.

Apply the product rule to $\frac{x^{\frac{1}{4}}}{3^{\frac{1}{4}}}$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{{(x^{\frac{1}{4}})}^{4}}{{(3^{\frac{1}{4}})}^{4}}) - 1 + 1}$

Simplify the numerator.

Multiply the exponents in ${(x^{\frac{1}{4}})}^{4}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x^{\frac{1}{4} \cdot 4}}{{(3^{\frac{1}{4}})}^{4}}) - 1 + 1}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x^{\frac{1}{4} \cdot 4}}{{(3^{\frac{1}{4}})}^{4}}) - 1 + 1}$

Rewrite the expression.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x^{1}}{{(3^{\frac{1}{4}})}^{4}}) - 1 + 1}$

Simplify.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x}{{(3^{\frac{1}{4}})}^{4}}) - 1 + 1}$

Simplify the denominator.

Multiply the exponents in ${(3^{\frac{1}{4}})}^{4}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x}{3^{\frac{1}{4} \cdot 4}}) - 1 + 1}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x}{3^{\frac{1}{4} \cdot 4}}) - 1 + 1}$

Rewrite the expression.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x}{3^{1}}) - 1 + 1}$

Evaluate the exponent.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x}{3}) - 1 + 1}$

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

Add $- 1$ and $1$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x}{3}) + 0}$

Add $\ln (\frac{x}{3})$ and $0$ .

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 e^{\ln (\frac{x}{3})}$

Exponentiation and log are inverse functions.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 (\frac{x}{3})$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = 3 (\frac{x}{3})$

Rewrite the expression.

$f (- \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}$ is the inverse of $f (x) = 3 e^{- 4 x + 1}$ .

$f^{- 1} (x) = - \frac{\ln (\frac{x}{3})}{4} + \frac{1}{4}$

Find the Inverse y=3e^(-4x+1)

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