Find the Inverse y = log of x-8

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Algebra

$y = \log (x - 8)$

Interchange the variables.

$x = \log (y - 8)$

Solve for $y$ .

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Rewrite the equation as $\log (y - 8) = x$ .

$\log (y - 8) = x$

Rewrite $\log (y - 8) = x$ 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 $b^{y} = x$ .

$10^{x} = y - 8$

Solve for $y$

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Rewrite the equation as $y - 8 = 10^{x}$ .

$y - 8 = 10^{x}$

Add $8$ to both sides of the equation.

$y = 10^{x} + 8$

Solve for $y$ and replace with $f^{- 1} (x)$ .

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

$f^{- 1} (x) = 10^{x} + 8$

Set up the composite result function.

$g (f (x))$

Evaluate $g (f (x))$ by substituting in the value of $f$ into $g$ .

$10^{\log (x - 8)} + 8$

Exponentiation and log are inverse functions.

$g (\log (x - 8)) = x - 8 + 8$

Combine the opposite terms in $x - 8 + 8$ .

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Add $- 8$ and $8$ .

$g (\log (x - 8)) = x + 0$

Add $x$ and $0$ .

$g (\log (x - 8)) = x$

Since $g (f (x)) = x$ , $f^{- 1} (x) = 10^{x} + 8$ is the inverse of $f (x) = \log (x - 8)$ .

$f^{- 1} (x) = 10^{x} + 8$

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