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May 15th, 2015, 05:12 PM   #1
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How do you simplify a two-variable division problem with variables in the exponent?

Simplify $\displaystyle \frac{x^{3a+2}} {x^{2a-1}}$

(In case I'm doing the LaTeX incorrectly, it's (x^3a+2) / (x^2a-1), or (x to the power of 3a + 2) divided by (x to the power of 2a - 1)

Since there's no answer to the problem, I wouldn't know how to set up a logarithm for either part of the fraction. (Sorry for how basic my knowledge is. I'm working on it.)

The answer is $\displaystyle x^{a + 3}$, but I can't see how to get there.

Last edited by skipjack; May 18th, 2015 at 11:16 PM.
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May 15th, 2015, 05:16 PM   #2
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Quote:
Originally Posted by Rexan View Post
Simplify $\displaystyle \frac{x^{3a+2}} {x^{2a-1}}$

Since there's no answer to the problem, I wouldn't know how to set up a logarithm for either part of the fraction. (Sorry for how basic my knowledge is. I'm working on it.)

The answer is $\displaystyle x^{a + 3}$, but I can't see how to get there.
Exponents have two properties that are useful here.

1) $\displaystyle \frac{1}{m^n} = m^{-n}$

2) $\displaystyle m^n \cdot m^p = m^{n + p}$

So
$\displaystyle \frac{x^{3a + 2}}{x^{2a - 1}} = \left ( x^{3a + 2} \right ) \cdot \left ( x^{-(2a - 1)} \right ) = x^{(3a + 2) - (2a - 1)}$

etc.

-Dan
Thanks from Rexan

Last edited by skipjack; May 18th, 2015 at 11:17 PM.
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May 15th, 2015, 05:24 PM   #3
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Quote:
Originally Posted by topsquark View Post
Exponents have two properties that are useful here.

1) $\displaystyle \frac{1}{m^n} = m^{-n}$

2) $\displaystyle m^n \cdot m^p = m^{n + p}$

So
$\displaystyle \frac{x^{3a + 2}}{x^{2a - 1}} = \left ( x^{3a + 2} \right ) \cdot \left ( x^{-(2a - 1)} \right ) = x^{(3a + 2) - (2a - 1)}$

etc.

-Dan
Finally got it. Thank you!
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