Find the Inverse y=(6^x-7)/4

$y = \frac{6^{x} - 7}{4}$

Step 1

Interchange the variables.

$x = \frac{6^{y} - 7}{4}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{6^{y} - 7}{4} = x$ .

$\frac{6^{y} - 7}{4} = x$

Multiply both sides of the equation by $4$ .

$4 \frac{6^{y} - 7}{4} = 4 x$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$4 \frac{6^{y} - 7}{4} = 4 x$

Rewrite the expression.

$6^{y} - 7 = 4 x$

Add $7$ to both sides of the equation.

$6^{y} = 4 x + 7$

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

$\ln (6^{y}) = \ln (4 x + 7)$

Expand $\ln (6^{y})$ by moving $y$ outside the logarithm.

$y \ln (6) = \ln (4 x + 7)$

Divide each term in $y \ln (6) = \ln (4 x + 7)$ by $\ln (6)$ and simplify.

Divide each term in $y \ln (6) = \ln (4 x + 7)$ by $\ln (6)$ .

$\frac{y \ln (6)}{\ln (6)} = \frac{\ln (4 x + 7)}{\ln (6)}$

Simplify the left side.

Cancel the common factor of $\ln (6)$ .

Cancel the common factor.

$\frac{y \ln (6)}{\ln (6)} = \frac{\ln (4 x + 7)}{\ln (6)}$

Divide $y$ by $1$ .

$y = \frac{\ln (4 x + 7)}{\ln (6)}$

Step 3

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

$f^{- 1} (x) = \frac{\ln (4 x + 7)}{\ln (6)}$

Step 4

Verify if $f^{- 1} (x) = \frac{\ln (4 x + 7)}{\ln (6)}$ is the inverse of $f (x) = \frac{6^{x} - 7}{4}$ .

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 $f^{- 1} (\frac{6^{x} - 7}{4})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{6^{x} - 7}{4}) = \frac{\ln (4 (\frac{6^{x} - 7}{4}) + 7)}{\ln (6)}$

Simplify the numerator.

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (\frac{6^{x} - 7}{4}) = \frac{\ln (4 (\frac{6^{x} - 7}{4}) + 7)}{\ln (6)}$

Rewrite the expression.

$f^{- 1} (\frac{6^{x} - 7}{4}) = \frac{\ln (6^{x} - 7 + 7)}{\ln (6)}$

Combine the opposite terms in $6^{x} - 7 + 7$ .

Add $- 7$ and $7$ .

$f^{- 1} (\frac{6^{x} - 7}{4}) = \frac{\ln (6^{x} + 0)}{\ln (6)}$

Add $6^{x}$ and $0$ .

$f^{- 1} (\frac{6^{x} - 7}{4}) = \frac{\ln (6^{x})}{\ln (6)}$

Expand $\ln (6^{x})$ by moving $x$ outside the logarithm.

$f^{- 1} (\frac{6^{x} - 7}{4}) = \frac{x \ln (6)}{\ln (6)}$

Cancel the common factor of $\ln (6)$ .

Cancel the common factor.

$f^{- 1} (\frac{6^{x} - 7}{4}) = \frac{x \ln (6)}{\ln (6)}$

Divide $x$ by $1$ .

$f^{- 1} (\frac{6^{x} - 7}{4}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\ln (4 x + 7)}{\ln (6)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (4 x + 7)}{\ln (6)}) = \frac{6^{\frac{\ln (4 x + 7)}{\ln (6)}} - 7}{4}$

Simplify the numerator.

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

$f (\frac{\ln (4 x + 7)}{\ln (6)}) = \frac{6^{\log_{6} (4 x + 7)} - 7}{4}$

Exponentiation and log are inverse functions.

$f (\frac{\ln (4 x + 7)}{\ln (6)}) = \frac{4 x + 7 - 7}{4}$

Subtract $7$ from $7$ .

$f (\frac{\ln (4 x + 7)}{\ln (6)}) = \frac{4 x + 0}{4}$

Add $4 x$ and $0$ .

$f (\frac{\ln (4 x + 7)}{\ln (6)}) = \frac{4 x}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{\ln (4 x + 7)}{\ln (6)}) = \frac{4 x}{4}$

Divide $x$ by $1$ .

$f (\frac{\ln (4 x + 7)}{\ln (6)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (4 x + 7)}{\ln (6)}$ is the inverse of $f (x) = \frac{6^{x} - 7}{4}$ .

$f^{- 1} (x) = \frac{\ln (4 x + 7)}{\ln (6)}$

Find the Inverse y=(6^x-7)/4

Related Questions

Support LCMGCF.com by disabling your adblocker.