# kinda hard exponent problem

• Sep 16th 2011, 08:33 AM
kinda hard exponent problem

Given 2x^2 = 3y^3

By how much will y increase if x is multiplied by 2?

1) 2^(1/3)
2) 4^(1/3)
3) 8

Attachment 22287

• Sep 16th 2011, 09:38 AM
Siron
Re: kinda hard exponent problem
Reform the given equation to $y=...$
Afterwards replace $x$ by $2x$.
Then you'll see $y$ is increased by a factor ... ?
• Sep 16th 2011, 09:44 AM
Re: kinda hard exponent problem
y= (2/3x^2)^(1/3)
2^2=4
4*2/3 = 8/3
8/3^(1/3)
that 3 keeps bugging me out!
• Sep 16th 2011, 09:50 AM
Siron
Re: kinda hard exponent problem
What I mean is the following:
$2x^2=3y^3$
$\Leftrightarrow y=\sqrt[3]{\frac{2x^2}{3}}$ (1)
Because the question asks how y increases if x is multiplied by two you have to replace x by 2x
$\Leftrightarrow y=\sqrt[3]{\frac{2(2x)^2}{3}}$
$\Leftrightarrow y=\sqrt[3]{\frac{2\cdot 4x^2}{3}}$
$\Leftrightarrow y=\sqrt[3]{4}\cdot \sqrt[3]{\frac{2x^2}{3}}$ (2)

So you recognize an equal part in (1) and (2), but (2) is multiplied by a factor ... ?
• Sep 16th 2011, 12:39 PM
Re: kinda hard exponent problem
I thought I read The problem wrong. That's why I included the image. Again I don't know the answer. I got the same reasoning solution as you
• Sep 16th 2011, 12:41 PM
e^(i*pi)
Re: kinda hard exponent problem
Quote:

If you understand how Siron got that answer then use your laws of exponents to say that $\sqrt[3]{4} = 4^{1/3}$
Edit: The question is asking to find the difference in "scale" if you will. In other words it wants you to find $\dfrac{f(2x)}{f(x)}$ where f(x) is the equation given.