If
$\displaystyle
x =\dfrac{e^{2x}}{y^2}
$
express z as a function on x and y
hmm, I'm a little rusty on my log laws as well. Basically I have an interview next week where I have to do a maths test. I have't used maths in 10 years, but in my time I was quite good at maths, Im afraid Ive forgotten a lot of the principles now. I would really appreciate it.
I'd have taken the previous answerer in base e so that you get
$\displaystyle \ln({xy^2}) = \ln{e^{2z}}$
Recalling that $\displaystyle \ln({e^a}) = a$ we get
$\displaystyle \ln{xy^2} = {2z}$ and then divide through by 2 to give:
$\displaystyle {z} = \frac{1}{2}\ln({xy^2})$