Im a noob mathematician.
I am trying to find the inverse of this function: $$F(x) = \frac{x + 299 \cdot 2^\frac{x-1}{7}}{4} $$
In the same way that: $$f(y) = ½(y - 1)$$ is the inverse of: $$f(x) = 2x + 1$$
I know the procedure goes along like this:
1) Let y = f(x)
2) Swap the x"s and y"s
3) Solve with y as the subject
The problem im having is with the exponent: (x-1)/7. I'm not sure what to do with it so that I can solve for y.
If someone could solve this for me that would be great. I am also interested in the steps to the solution, so including the steps would be a bonus.
Thanks in advance!
In general the equation,
$$e^{-cx}=a_0\left(x-r \right)$$
Has solution,
$$r+\frac{1}{c} W(\frac{ce^{-cr}}{a_0})=x$$
The reason is below this line.
Using the lambert W function. Let $x-r=u$ so that $x=r+u$.
$$e^{-c(r+u)}=a_0 u $$
$$e^{-cr}e^{-cu}=a_0u$$
$$e^{-cu}=a_0e^{cr} u$$
$$1=a_0e^{cr} u e^{cu}$$
$$\frac{e^{-cr}}{a_0}=u e^{cu}$$
$$\frac{ce^{-cr}}{a_0}=cu e^{cu}$$
$$W(\frac{ce^{-cr}}{a_0})=cu$$
$$\frac{1}{c} W(\frac{ce^{-cr}}{a_0})=u$$
$$r+\frac{1}{c} W(\frac{ce^{-cr}}{a_0})=x$$
The equation you have written can be rearranged to the form considered previously.
$$4F=x+299(2^{-\frac{1}{7}}) e^{x \frac{\ln 2}{7}}$$
$$\frac{4F-x}{299}=2^{-\frac{1}{7}} e^{x \frac{\ln 2}{7}}$$
$$\frac{2^{\frac{1}{7}}}{299} \left( 4F-x \right)=e^{ x \frac{\ln 2}{7}}$$
$$\frac{-2^{\frac{1}{7}}}{299} \left( x-4F \right)=e^{ x \frac{\ln 2}{7}}$$
Thus, to find the inverse use $r=4x$, $a_0= \frac{-2^{\frac{1}{7}}}{299}$, and $c=-\frac{\ln 2}{7}$.