Given that $\displaystyle \frac{(144p^4)^{\frac{3}{2}}}{(216p^{-3})^\frac{-2}{3}}=2^x3^yp^z$
Find the values of x, y and z
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
$\displaystyle \dfrac{(144p^4)^{\frac{3}{2}}}{(216p^{-3})^\frac{-2}{3}}=2^x3^yp^z$
$\displaystyle (144p^4)^\frac{3}{2} \times (216p^{-3})^\frac{2}{3}}=2^x3^yp^z$
$\displaystyle 1728p^6 \times 36p^{-2}=2^x3^yp^z$
$\displaystyle 62208p^4 =2^x3^yp^z$
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?
The questions wants me to find the value of x, y and z
answer is x=8, y=5, z=4
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...
That is a badly worded question.
There are infinite solutions to what it is asking, for example:
p=2, y=1, z=1, x=$\displaystyle \frac{\log{165888}}{\log{2}}$ and this would still satisfy the question. This was just from me substituting random digits for 3 variables and solving for the fourth.
What numbers?
As it is, your problem is ambiguous; please submit the original in full.
Or ask your teacher what the purpose is.
Since your equation can be rearranged this way: 2^x 3^y / p^(4-z) = 62208,
then we can set p=1 and z=3, so that we're dealing with 2^x 3^y = 62208