What is the inverse of $y = log (x - 4) + 2$ $y=log(x-4)+2$ ?

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Answer 1 · 2018-03-02T11:18:31

$10^{x - 2} + 4$ $10^(x-2)+4$ is the inverse.

We have the function $f (x) = y = log (x - 4) + 2$ $f(x)=y=log(x-4)+2$

To find $f^{- 1} (x)$ $f^-1(x)$ , we take our equation:

$y = log (x - 4) + 2$ $y=log(x-4)+2$

Switch the variables:

$x = log (y - 4) + 2$ $x=log(y-4)+2$

And solve for $y$ $y$ :

$x - 2 = log (y - 4)$ $x-2=log(y-4)$

We can write $x - 2$ $x-2$ as $log (10^{x - 2})$ $log(10^(x-2))$ , so we have:

$log (10^{x - 2}) = log (y - 4)$ $log(10^(x-2))=log(y-4)$

As the bases are the same:

$y - 4 = 10^{x - 2}$ $y-4=10^(x-2)$

$y = 10^{x - 2} + 4$ $y=10^(x-2)+4$

Which is your inverse.

What is the inverse of y=log(x-4)+2y=log(x−4)+2y=log(x-4)+2 ?