Find the Inverse f(x)=7 log of 1/4x-3+2

Rewrite the equation as ${\displaystyle 7\log (\frac{1}{4}y-3)+2=x}$.

Subtract ${\displaystyle 2}$ from both sides of the equation.

Divide each term in ${\displaystyle 7\log (\frac{1}{4}y-3)=x-2}$ by ${\displaystyle 7}$ and simplify.

Rewrite ${\displaystyle \log (\frac{1}{4}y-3)=\frac{x}{7}-\frac{2}{7}}$ in exponential form using the definition of a logarithm. If ${\displaystyle x}$ and ${\displaystyle b}$ are positive real numbers and ${\displaystyle b\ne 1}$, then ${\displaystyle {\log }_b\left(x\right)=y}$ is equivalent to ${\displaystyle b^y=x}$.

Solve for ${\displaystyle y}$.

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

Evaluate ${\displaystyle f^{-1}\left(f\left(x\right)\right)}$.

Evaluate ${\displaystyle f\left(f^{-1}\left(x\right)\right)}$.

Since ${\displaystyle f^{-1}\left(f\left(x\right)\right)=x}$ and ${\displaystyle f\left(f^{-1}\left(x\right)\right)=x}$, then ${\displaystyle f^{-1}\left(x\right)=4\cdot {10}^{\frac{x}{7}-\frac{2}{7}}+12}$ is the inverse of ${\displaystyle f\left(x\right)=7\log (\frac{1}{4}x-3)+2}$.