1. ## Functions

If

$\displaystyle x =\dfrac{e^{2x}}{y^2}$

express z as a function on x and y

2. what is the relation between x,y and z?

3. Im afraid thats all that is in the question! Nothing else - this is a from a sample maths exam that is for interviewing purposes.

4. there is something missing in your question
I think it is y^z
not y^2

5. Sorry it was indeed a typo in my original post!

$\displaystyle x =\dfrac{e^{2z}}{y^2}$

6. Originally Posted by tangehayes
Sorry it was indeed a typo in my original post!

$\displaystyle x =\dfrac{e^{2z}}{y^2}$
Now, it should be easy putting x,y on the same side and using the logarithmic function right!

7. $\displaystyle x =\dfrac{e^{2z}}{y^2}$
then
$\displaystyle x y^2=e^{2z}$

and we know if a=b, then, log a = log b

$\displaystyle log (x y^2)=log (e^{2z})$

can you complete it now by using the logarithmic basic`s laws ?

8. 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.

9. Originally Posted by tangehayes
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:

$\displaystyle {z} = \frac{1}{2}\ln({xy^2})$