Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Try multiplying by $y^h/y^h$ to clear the denominator downstairs. – Brett Frankel Jan 29 '13 at 16:30
up vote 3 down vote accepted

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

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

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

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

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

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


share|cite|improve this answer
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.

share|cite|improve this answer
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 $x = (x^h)^{\large \frac 1h}$:

$$ \begin{align} k &= \frac{1}{1+\left(\large\frac{x}{y}\right)^h}= \frac{1}{1+\left(\large\frac{x^h}{y^h}\right)}\cdot \frac{y^h}{y^h} \\ \\ \\ k&=\frac{y^h}{y^h+x^h} \\ \\ \\ k(y^h + x^h) &= y^h \\ \\ \\ y^h+x^h &= y^h/k\\ \\ \\ x^h &= \left(\frac{y^h}{k} - y^h\right) = y^h\left(\frac 1k - 1\right)\\ \\ \\ x &= y\left(\frac 1k - 1\right)^{\large \frac 1h} \end{align} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.