## GRE practice with fractions and exponents

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
wchriste
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### GRE practice with fractions and exponents

I've recently begun studying for the GRE, and this was the problem:
(1/2)^a=(1/3)^b

These were the answer choices (I'm not sure if they are important for this, so I'm including them just in case):
Which is true?
a<b<0
b>a>0
b<a<0
b<0<a
a<0<b

The solution to this problem began with:
"From (1/2)^a=(1/3)^b, we know that 2^a=3^b"

However, I'm not sure how "we know" this. I tried setting a=2 and seeing if (1/2)^a and 2^a were the same, but I got 1/4 and 4. Any clarification would be greatly appreciated, as questions like these seem to come up a lot in practice and if there are assumptions I can make, I want to know them Thanks!

QM deFuturo
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### Re: GRE practice with fractions and exponents

You can review the basic exponent laws here :

http://www.purplemath.com/modules/exponent.htm

But basically, when you have (x/y)^n, you can write this as (x)^n/(y)^n.

If you do that, and remember what 1^n equals, you may figure out how to rearrange the equation in a form that makes it easier for you to compare the two sides.

QM

stapel_eliz
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To expand upon the (excellent) previous reply, you are given that:

. . . . .$\left(\frac{1}{2}\right)^a\, =\, \left(\frac{1}{3}\right)^b$

. . . . .$\frac{1^a}{2^a}\, =\, \frac{1^b}{3^b}$

. . . . .$\frac{1}{2^a}\, =\, \frac{1}{3^b}$

The only way that these fractions can be equal is for their denominators to be equal, since we can see that their numerators are the same. In other words:

. . . . .$2^a\, =\, 3^b$

Since 2 < 3 and since no answer-option allows for $a\, =\, b$ (so, in particular, we can not have a = b = 0), then the powers must be different. If the powers are positive, then what must be true of the relative sizes of $a$ and $b$? If the powers are negative, so we're looking at inverted fractions with the absolute values of the powers, what then must be true of their relative sizes? Is there any way in which oppositely signed powers could fulfill the given equality?

michaelempeigne
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### Re: GRE practice with fractions and exponents

$(\frac{1} {2})^a = (\frac{1}{3})^b$
$2^a = 3^b$
$a ln 2 = b ln 3$
$a / b = ln 3 / ln 2$
so $b < a >0$

stapel_eliz
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so $b < a >0$
I'm sorry, but I don't understand what the above is meant to say...?

wchriste
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### Re: GRE practice with fractions and exponents

All of these answers are very helpful, and I understand why the problem is solved that way. The GRE doesn't require using logarithms, but nothing stops me from doing so (and I seem to remember those better than exponents -_-) and it never hurts to know a little extra.

Thank you all so much!