1. ## Algebra

Given that $\frac{(144p^4)^{\frac{3}{2}}}{(216p^{-3})^\frac{-2}{3}}=2^x3^yp^z$

Find the values of x, y and z

2. I'm not getting the question.

There are 4 variables in 1 equation, and you want to find the values of 3 of them? There would be infinite solutions.
So I'm assuming you want to rewrite the equation in the form x = ... y = ... and z = ... ?

First, we can simplify the equation

$\dfrac{(144p^4)^{\frac{3}{2}}}{(216p^{-3})^\frac{-2}{3}}=2^x3^yp^z$

$(144p^4)^\frac{3}{2} \times (216p^{-3})^\frac{2}{3}}=2^x3^yp^z$

$1728p^6 \times 36p^{-2}=2^x3^yp^z$

$62208p^4 =2^x3^yp^z$

Now can you use the rules of logarithms to rewrite them in the form x=... and so on?
Or are you asking for something else?

3. The questions wants me to find the value of x, y and z

I have done wat u have done before i started this thread, couldnt find a way to change that number into powers of 2 and 3. Z was easily done problem lies in x and y for me...

4. That is a badly worded question.

There are infinite solutions to what it is asking, for example:
p=2, y=1, z=1, x= $\frac{\log{165888}}{\log{2}}$ and this would still satisfy the question. This was just from me substituting random digits for 3 variables and solving for the fourth.

5. Sorry the actual question should be:

[QUOTE=Punch;562593]Given that $\frac{(144p^4)^{\frac{3}{2}}}{(216p^{-3})^\frac{-2}{3}}=2^x3^yp^z$

Evaluate x,y and z

6. You can write 144 and 216 as

$144 = 2^43^2 and 216 = 2^33^3$

Substitute in the given problem and simplify and compare the powers of 2, 3 and p.

7. Originally Posted by Educated
That is a badly worded question.
Agree.
You can even solve for p: p = (2^x 3^y / 62208)^[1 / (4-z)]

8. Hi. thanks for your reply, can u show me the process of getting those numbers?

9. Originally Posted by Punch
Hi. thanks for your reply, can u show me the process of getting those numbers?
What numbers?
As it is, your problem is ambiguous; please submit the original in full.