Proportionality

**Jen1603** · October 21st 2008, 05:21 AM

1. The quantity Q is proportional to x^3 and inversely proportional to
(y+a)^4 . Find a formula for Q given Q=9 when x=3 and y=0.

2. Also :

Express (32/27) / 18 x (square root of) 3.

in the form 2^a x 3^b

thankkkks!

**ajj86** · October 21st 2008, 06:03 AM

Hopefully I am reading this correctly, but I would start by factoring the numbers and then separating them.

32 = 2^5
27 = 3^3
18 = 2 * 3^2
sqrt(3) = 3^1/2

So, the original form can be rewritten as:

(32/2) * 1/(3^3*3^2*3^(1/2))

which can be reduced to

16/(3^(11/2)) or 16/(3^5.5)

The exponent of 5.5 is found by adding 3 + 2 + 0.5 since multiplying numbers with the same base, which in this case is 3, and different exponents requires that they be added.

Now, our task is to express it in the form of 2^a * 3^b.

16 = 2^4

So, the fraction looks like:

(2^4)/(3^5.5)

Since 3^5.5 is in the denominator, the exponent becomes negative, and the final result looks like:

2^4 * 3^-5.5 or 2^4 * 3^(-11/2)

I hope this is helpful.

**ajj86** · October 21st 2008, 06:20 AM

If Q is proportional to x^3 and also Q is proportional to 1/(y+a)^4, then we could set up an equation along the lines of

Q = x^3 / (y+a)^4

We are given initial values:

Q = 9, x = 3 and y = 0.

Substitute these to yield:

9 = 3^3 / (0+a)^4

Reduce: 9 = 27/a^4

Clear the fractions by multiplying both sides by a^4 and dividing by 9 to get:

a^4 = 3
Solve for a: a = 3^(1/4) or 3^0.25

Now, rewrite the original equation

Q = x^3 / (y + 3^0.25)^4

**Moo** · October 22nd 2008, 11:32 AM

Hello,

Originally Posted by Jen1603

1. The quantity Q is proportional to x^3 and inversely proportional to
(y+a)^4 .

"Q is proportional to x^3" means that there exists a number k such that $Q=k*x^3$
"Q is inversely proportional to (y+a)^4" means that there exists a number m such that $Q=\frac{m}{(y+a)^4}$

Find a formula for Q given Q=9 when x=3 and y=0.

When $x=3$ , we get from the first formula, $9=Q=k*(3)^3 \implies k=\frac{9}{3^3}=\frac 13$

When $y=0$ , we get from the second formula, $9=Q=\frac{m}{a^4} \implies m=9a^4$

So there are two formulae for Q (by substituting k and m) :
$Q=\frac{x^3}{3}=\frac{9a^4}{(y+a)^4}$

**Shyam** · October 22nd 2008, 12:31 PM

= $\frac{32}{27} \times \frac{1}{18}\times \sqrt{3}$

$= \frac{32}{27 \times 18} \times 3^{\frac{1}{2}}$

$= \frac{2^5}{3^3 \times 2 \times 3^2} \times 3^{\frac{1}{2}}$

$<br /> = \frac{2^4}{3^5} \times 3^{\frac{1}{2}}$

$=2^4 \times 3^{(\frac{1}{2}-5)}$

$=2^4 \times 3^{\frac{-9}{2}}$

**batman** · October 22nd 2008, 01:01 PM

Note that 32=2^5, 27=3^3, 18=2*3^2, sqrt(3)=3^(-0.5).
Also note that y/x^n=y*x^-n and x^n*x^m=x^(n+m).

Can you solve it now?

**Jen1603** · October 22nd 2008, 01:04 PM

I got that far, im just not sure what to do with the 32/27 on top of the other fraction ?

**batman** · October 22nd 2008, 01:11 PM

Apply the rule y^m/x^n=y^m*x^-n twice.
(32/27) / 18 = 2^5*3^(-3) / (2*3^2) = 2^5*3^(-3)*(2^(-1)*3^(-2)) = 2^4 * 3^(-5).

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