How do you find the inverse of $g (x) = {log}_{3} (x + 2) - 6$ $g(x)= log_3(x+2)-6$ ?

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Answer 1 · 2015-12-03T15:56:02

$g^{- 1} (x) = 3^{x + 6} - 2$ $g^-1(x)=3^(x+6)-2$

Write as:

$\times y = {log}_{3} (z + 2) - 6$ $color(white)(xx)y=log_3(z+2)-6$

Switch $x$ $x$ and $y$ $y$ .

$\times x = {log}_{3} (y + 2) - 6$ $color(white)(xx)x=log_3(y+2)-6$

Solve for $y$ $y$ .

$\times x + 6 = {log}_{3} (y + 2)$ $color(white)(xx)x+6=log_3(y+2)$

$\times 3^{x + 6} = 3^{{log}_{3} (y + 2)}$ $color(white)(xx)3^(x+6)=3^(log_3(y+2))$

$\times 3^{x + 6} = y + 2$ $color(white)(xx)3^(x+6)=y+2$

$\times 3^{x + 6} - 2 = y$ $color(white)(xx)3^(x+6)-2=y$

Rewrite with $g^{- 1} (x)$ $g^-1(x)$ instead of $y$ $y$ .

$\times g^{- 1} (x) = 3^{x + 6} - 2$ $color(white)(xx)g^-1(x)=3^(x+6)-2$

How do you find the inverse of g(x)= log_3(x+2)-6g(x)=log3(x+2)−6g(x)= log_3(x+2)-6?