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by **maroonblazer** on Sun Dec 04, 2011 6:49 pm

Hi again,
The problem states:
If $(\frac{3}{4})^{-3} + (\frac{8}{3})^2 = \frac{2^a}{3^b}$
find $a$ and $b$

The answer is a=8 and b=3, but I'm stumped as to how they got there.

It seems like I need to turn the right-hand side into an addition expression. I can convert it to a product by expressing $3^b$ as $3^{-b}$ , but I don't know what to do next to get it to an addition expression. Perhaps that's not the right strategy?

Thanks in advance,
mb

**Sponsor**

by **stapel_eliz** on Sun Dec 04, 2011 7:14 pm

maroonblazer wrote:If $(\frac{3}{4})^{-3} + (\frac{8}{3})^2 = \frac{2^a}{3^b}$
find $a$ and $b$

The answer is a=8 and b=3, but I'm stumped as to how they got there.

When in doubt, do what you can with what you have, and see where that leads. With luck, you'll start seeing something helpful.

. . . . . $\left(\frac{3}{4}\right)^{-3}\, +\, \left(\frac{8}{3}\right)^2\, =\, \frac{2^a}{3^b}$

. . . . . $\left(\frac{4}{3}\right)^3\, +\, \left(\frac{2^3}{3}\right)^2\, =\, \frac{2^a}{3^b}$

. . . . . $\left(\frac{2^2}{3}\right)^3\, +\, \frac{(2^3)^2}{3^2}\, =\, \frac{2^a}{3^b}$

. . . . . $\frac{(2^2)^3}{3^3}\, +\, \frac{2^{3\times 2}}{3^2}\, =\, \frac{2^a}{3^b}$

Do you see where to go now? :wink:

by **maroonblazer** on Sun Dec 04, 2011 8:07 pm

Do you see where to go now?

Ahh...so:
$\frac{2^6}{3^3} + \frac{2^6}{3^2} = \frac{2^a}{3^b}$

$\frac{2^6}{3^3} + (\frac{2^6}{3^2}*\frac{3}{3}) = \frac{2^a}{3^b}$

$\frac{2^6}{3^3} + \frac{3*2^6}{3^3} = \frac{2^a}{3^b}$

$\frac{64}{3^3} + \frac{192}{3^3} = \frac{2^a}{3^b}$

$\frac{256}{3^3} = \frac{2^a}{3^b}$

$\frac{2^8}{3^3} = \frac{2^a}{3^b}$ :clap:

I have to say though:

When in doubt, do what you can with what you have, and see where that leads. With luck, you'll start seeing something helpful.

I find this to be one of the most frustrating things about math. It seems like so much of it is rooting around aimlessly in the hope of stumbling upon something. It seems like, for being an "exact" science, it should be more exact.

As always I really appreciate your help!
mb

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