Basic algebra inversion I need to solve the following equation for $x$:
$$k= \frac{1}{1+(\frac{x}{y})^h}$$
Does anybody know the step-by-step process for rearranging this?
I think I need to invert the function, but I'm not sure about how to handle the ^$h$.
Thank you for any help.
 A: $$k= \frac{1}{1+(\frac{x}{y})^h}$$
$$k (1+(\frac{x}{y})^h) = 1$$
$$(1+(\frac{x}{y})^h) = \frac{1}{k}$$
$$(\frac{x}{y})^h = (\frac{1}{k} - 1)$$
$$(\frac{x}{y}) = (\frac{1}{k} - 1)^{\frac{1}{h}}$$
$$x = y(\frac{1}{k} - 1)^{\frac{1}{h}}$$
Regards
A: I multiply the Right-hand side by $\dfrac{y^h}{y^h}$ to clear the fraction in the denominator, then simply work to isolate $x^h$, and then solve for $x = (x^h)^{\large \frac 1h}$:
$$
\begin{align}
k &= \frac{1}{1+\left(\large\frac{x}{y}\right)^h}= \frac{1}{1+\left(\large\frac{x^h}{y^h}\right)}\cdot \frac{y^h}{y^h} \\ \\ \\
k&=\frac{y^h}{y^h+x^h} \\ \\ \\
k(y^h + x^h) &= y^h \\ \\ \\
y^h+x^h &= y^h/k\\ \\ \\
x^h &= \left(\frac{y^h}{k} - y^h\right) = y^h\left(\frac 1k - 1\right)\\ \\ \\
x &= y\left(\frac 1k - 1\right)^{\large \frac 1h}
\end{align}
$$
A: Ask yourself:  if you were given values for the variables on the RHS and asked to calculate k, how would you do it? You'd divide x by y, raise the result to the h-th power, add 1,  and find the inverse.
You need to undo each step in the calculation, in order, to work your way down to $x=$
So: invert both sides;  subtract 1 from both sides;  take the h-th root of both sides;  and multiply both sides by y.
