# 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.

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Try multiplying by $y^h/y^h$ to clear the denominator downstairs. – Brett Frankel Jan 29 '13 at 16:30

$$k= \frac{1}{1+(\frac{x}{y})^h}$$

$$k (1+(\frac{x}{y})^h) = 1$$

$$(1+(\frac{x}{y})^h) = \frac{1}{k}$$

$$(\frac{x}{y})^h = (\frac{1}{k} - 1)$$

$$(\frac{x}{y}) = (\frac{1}{k} - 1)^{\frac{1}{h}}$$

$$x = y(\frac{1}{k} - 1)^{\frac{1}{h}}$$

Regards

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And Regards to you, my friend $\Large !$ – amWhy May 5 '13 at 0:21

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.

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thanks for very much for the help, this is much appreciated. I understand the solution to the problem now :) – micronaut Jan 29 '13 at 20:56

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}

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