# Thread: inverse function

1. ## inverse function

Hi i'm having problem finding inverse of this exponential equation.

f(x)=4-2^(-x+1)

f(y)=4-2^(-y+1)

solve for y:

x=4-2^-y+1

x= 4- (1/2^(-y+1))

What do i do next?

2. Originally Posted by Anemori
Hi i'm having problem finding inverse of this exponential equation.

f(x)=4-2^(-x+1)

f(y)=4-2^(-y+1)

solve for y:

x=4-2^-y+1

x= 4- (1/2^(-y+1))

What do i do next?
$x=4-2^{-y+1}$

$\Rightarrow 4-x=2^{-y+1}$

$\Rightarrow \log{(4-x)}=\log{2^{-y+1}}$

$\Rightarrow \log{(4-x)}=(-y+1)\log{2}$

$\Rightarrow \frac{\log{(4-x)}}{\log 2}=-y+1$

$\Rightarrow y=1-\frac{\log{(4-x)}}{\log 2}$

3. Originally Posted by Anemori
Hi i'm having problem finding inverse of this exponential equation.

f(x)=4-2^(-x+1)

f(y)=4-2^(-y+1)

solve for y:

x=4-2^-y+1

x= 4- (1/2^(-y+1))

What do i do next?
An equivalent solution to the one posted above:

$x = 4 - 2^{1 - y}$

$4 - x = 2^{1 - y}$

$log_2(4 - x) = 1 - y$

$y = 1 - log_2(4 - x)$

$f^{-1}(x) = 1 - log_2(4 - x)$

If you think about it you should see that in general, $log_b(b^u) = u$, for suitable choice of b.