# Thread: Make x the subject

1. ## Make x the subject

Make x the subject of the formula: $v=\frac{1}{k}\sqrt{[1+(\frac{1}{x}+1)^2}/Lg$ In case I made a mistake in the coding, its v=1/k sqrt(1+(1/x +1)^2/Lg)

2. $\text{Lg}$ as in the logarithm with base 2? The formula doesn't make sense (i.e. the denominator).

3. No Lg are just two different letters. The second part of the question shows this when they ask you to put in values for each. k = 1.26, L = 48.55, g = 32.21, v = 0.0483 and solve for x.

4. So we have $v = \frac{\sqrt{\left[1 + \left(\frac{1}{x}+1 \right)^{2} \right]}}{kLg}$

Then $vkLg = \sqrt{\left[1 + \left(\frac{1}{x}+1 \right)^{2} \right]}$

Or $v^{2}k^{2}L^{2}g^{2}-1 = \left(\frac{1}{x} +1 \right)^2$

So $\left(\frac{1}{x} +1 \right)^2 = \frac{1}{x^2} + \frac{2}{x} +1$ and $v^{2}k^{2}L^{2}g^{2}-2 = \frac{1}{x^2} + \frac{2}{x}$

5. Originally Posted by tukeywilliams
So we have $v = \frac{\sqrt{\left[1 + \left(\frac{1}{x}+1 \right)^{2} \right]}}{kLg}$

Then $vkLg = \sqrt{\left[1 + \left(\frac{1}{x}+1 \right)^{2} \right]}$

Or $v^{2}k^{2}L^{2}g^{2}-1 = \left(\frac{1}{x} +1 \right)^2$
Actually it's simpler at this point to use:
$\frac{1}{x} + 1 = \pm \sqrt{v^{2}k^{2}L^{2}g^{2}-1}$

and solve for x from there.

-Dan