q23 (answer 4)

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

thanks in advance!

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- Sep 16th 2011, 07:33 AMjsvladkinda hard exponent problem
q23 (answer 4)

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

thanks in advance! - Sep 16th 2011, 08:38 AMSironRe: kinda hard exponent problem
Reform the given equation to $\displaystyle y=...$

Afterwards replace $\displaystyle x$ by $\displaystyle 2x$.

Then you'll see $\displaystyle y$ is increased by a factor ... ? - Sep 16th 2011, 08:44 AMjsvladRe: kinda hard exponent problem
y= (2/3x^2)^(1/3)

2^2=4

4*2/3 = 8/3

8/3^(1/3)

isn't an answer option... grrr

that 3 keeps bugging me out! - Sep 16th 2011, 08:50 AMSironRe: kinda hard exponent problem
What I mean is the following:

$\displaystyle 2x^2=3y^3$

$\displaystyle \Leftrightarrow y=\sqrt[3]{\frac{2x^2}{3}}$ (1)

Because the question asks how y increases if x is multiplied by two you have to replace x by 2x

$\displaystyle \Leftrightarrow y=\sqrt[3]{\frac{2(2x)^2}{3}}$

$\displaystyle \Leftrightarrow y=\sqrt[3]{\frac{2\cdot 4x^2}{3}}$

$\displaystyle \Leftrightarrow y=\sqrt[3]{4}\cdot \sqrt[3]{\frac{2x^2}{3}}$ (2)

So you recognize an equal part in (1) and (2), but (2) is multiplied by a factor ... ? - Sep 16th 2011, 11:39 AMjsvladRe: 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, 11:41 AMe^(i*pi)Re: kinda hard exponent problem
If you understand how Siron got that answer then use your laws of exponents to say that $\displaystyle \sqrt[3]{4} = 4^{1/3}$

Edit: The question is asking to find the difference in "scale" if you will. In other words it wants you to find $\displaystyle \dfrac{f(2x)}{f(x)}$ where f(x) is the equation given.

From Siron's post the answer is eq2/eq1