# Thread: Rational Exponents

1. ## Rational Exponents

1) Simplify:

I know I can split it up like below, but I'm finding it hard to remember where to go next.

It's obvious that both 9 and 15 are factors of 45, but, in terms of exponents, I can't remember the order of operations from here.

2) Simplify:

I know that a^(1/n) = nth root of a, but I can't seem to get the correct answer.

2. ## Re: Rational Exponents

More important than 9 and 15 being factors of 45 is that 3 and 5 are factors of all three numbers: $\displaystyle 9= 3^2$, 15= 3(5), and $\displaystyle 45= (3^2)(5)$. So [tex](45)^{1/3}= ((3^2)(5))^{1/3}= (3^{2/3})(5^{1/3}), $\displaystyle 9^{3/4}= (3^2)^{3/4}= 3^{3/2}$, and $\displaystyle 15^{3/2}= (5^{3/2})(3^{3/2})$.

So $\displaystyle \frac{45^{1/3}}{(9^{3/4})(15^{3/2})}= \frac{(3^{2/3})(5^{1/3})}{(3^{3/2})(3^{3/2})(5^{3/2})}$.

Can you do it now?

3. ## Re: Rational Exponents

Yes, thanks for the help.

Could someone point me in the right direction for 2?

4. ## Re: Rational Exponents

Originally Posted by Fratricide
Could someone point me in the right direction for 2?
$\displaystyle \frac{1}{\sqrt{x-1}}+\sqrt{x-1}=\frac{x}{\sqrt{x-1}}=\frac{x\sqrt{x-1}}{x-1}$

5. ## Re: Rational Exponents

If you are asking "How can I look at this problem and instantly know the answer without doing any calculation at all?", I can't help you because I certainly cannot do that. But in about the fourth or fifth grade I learned that to add fractions, you need to get a "common denominator". To add $\displaystyle \frac{1}{\sqrt{x- 1}}+ \sqrt{x-1}= \frac{1}{\sqrt{x- 1}}+ \frac{\sqrt{x-1}}{1}$, you need the "common denominator" which, obviously, is $\displaystyle \sqrt{x- 1}$. Multiplying both numerator and denominator of $\displaystyle \frac{\sqrt{x- 1}}{1}$ by $\displaystyle \sqrt{x- 1}$ gives
$\displaystyle \frac{1}{\sqrt{x- 1}}+ \frac{x- 1}{\sqrt{x- 1}}$

6. ## Re: Rational Exponents

I understand it up to there, but after that I'm not sure what to do.

That is, I can't grasp how Plato got from here:

Originally Posted by Plato
$\displaystyle \frac{x}{\sqrt{x-1}}$
To here:

Originally Posted by Plato
$\displaystyle \frac{x\sqrt{x-1}}{x-1}$

Note: The answer in the textbook is: x(x-1)^-1​.

7. ## Re: Rational Exponents

Originally Posted by Fratricide
I understand it up to there, but after that I'm not sure what to do.
That is, I can't grasp how Plato got from here:

To here:

Note: The answer in the textbook is: x(x-1)^-1​.
Note that answer is incorrect.

$\displaystyle \frac{x}{\sqrt{x-1}}$ can be written as $\displaystyle x(x-1)^{-1/2}$

But in general: $\displaystyle \frac{x}{\sqrt{x-1}}=\frac{x}{\sqrt{x-1}}\cdot\frac{\sqrt{x-1}}{\sqrt{x-1}}=\frac{x\sqrt{x-1}}{x-1}$

Have a look at this.