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?
$\displaystyle x=4-2^{-y+1}$
$\displaystyle \Rightarrow 4-x=2^{-y+1}$
$\displaystyle \Rightarrow \log{(4-x)}=\log{2^{-y+1}}$
$\displaystyle \Rightarrow \log{(4-x)}=(-y+1)\log{2}$
$\displaystyle \Rightarrow \frac{\log{(4-x)}}{\log 2}=-y+1$
$\displaystyle \Rightarrow y=1-\frac{\log{(4-x)}}{\log 2}$
An equivalent solution to the one posted above:
$\displaystyle x = 4 - 2^{1 - y}$
$\displaystyle 4 - x = 2^{1 - y}$
$\displaystyle log_2(4 - x) = 1 - y$
$\displaystyle y = 1 - log_2(4 - x)$
$\displaystyle f^{-1}(x) = 1 - log_2(4 - x)$
If you think about it you should see that in general, $\displaystyle log_b(b^u) = u$, for suitable choice of b.