1. ## Algebra and Functions

Hi I have just done my first day doing A levels. My teacher has given me homework to do with simplifying and evaluating expressions. I don't understand the method on how to solve the equation.

An example of one of the simplifying questions is: 5x^(1+2/5) / x^(2/5)

An example of one of the evaluation questions is: 27/8^(2/3)

My apologies for not using the LaTeX software but I tried for about half an hour to input the equation correctly but only got errors.

2. ## Re: Algebra and Functions

Originally Posted by jacobjacob44
Hi I have just done my first day doing A levels. My teacher has given me homework to do with simplifying and evaluating expressions. I don't understand the method on how to solve the equation.

An example of one of the simplifying questions is: 5x^(1+2/5) / x^(2/5)

An example of one of the evaluation questions is: 27/8^(2/3)

My apologies for not using the LaTeX software but I tried for about half an hour to input the equation correctly but only got errors.
The first one, use $\displaystyle \frac{a^m}{a^n} = a^{m - n}$.

The second one, $\displaystyle 8^{\frac{2}{3}} = \left(8^{\frac{1}{3}\right)^2 = \left(\sqrt[3]{8}\right)^2$...

3. ## Re: Algebra and Functions

what you will do when the problem is X^3/ X^2. same logic.

4. ## Re: Algebra and Functions

Thanks for the first one that has worked great.

As for the second one I have realised I made a mistake. I didnt mean for it to look like 27 divided by 8 ^ 2/3.
I meant for it to look like $(\frac{27}{8})^\frac{2}{3}$
Still learning on how to work the software.

5. ## Re: Algebra and Functions

Originally Posted by jacobjacob44
Thanks for the first one that has worked great.

As for the second one I have realised I made a mistake. I didnt mean for it to look like 27 divided by 8 ^ 2/3.
I meant for it to look like $(\frac{27}{8})^\frac{2}{3}$
Still learning on how to work the software.
You should know that $\displaystyle \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$...

6. ## Re: Algebra and Functions

I'll get you started:

$(\frac{27}{8})^{\frac{2}{3}}=\frac{27^{2/3}}{8^{2/3}}$

7. ## Re: Algebra and Functions

Okay, I understand the working behind what you have both said. What is confusing me however is in the notes I took from what my teacher did in class included square and cube roots e.g an example she gave us:
$(\frac{64}{81})^\frac{1}{2} = \sqrt{\frac{64}{81}} = \frac{\sqrt{64}}{\sqrt{81}} = \frac{8}{9}$
Sorry if this is something obvious that I am not getting, just its all very confusing for me at the moment

8. ## Re: Algebra and Functions

if a term is raised to the 1/2 power , that is equivalent to the square root of that term.

the square root of a fraction is equivalent to the square root of the numerator divided by the square root of the denominator.

9. ## Re: Algebra and Functions

Okay, so would the way to work out my original question $(\frac{27}{8})^\frac{2}{3}$ work out something like this?

$(\frac{27}{8})^\frac{2}{3} = \frac{27^{2/3}}{8^{2/3}} = \frac{3}{2}$

3 being the cube root of 27 and 2 being the cube root of 8?

10. ## Re: Algebra and Functions

Originally Posted by jacobjacob44
Okay, so would the way to work out my original question $(\frac{27}{8})^\frac{2}{3}$ work out something like this?

$(\frac{27}{8})^\frac{2}{3} = \frac{27^{2/3}}{8^{2/3}} = \frac{3}{2}$

3 being the cube root of 27 and 2 being the cube root of 8?
Almost there. Now you need to square each term, since it's to the power of 2/3, not 1/3.

11. ## Re: Algebra and Functions

Ahh I see! So would that make the final answer be $\frac{9}{6}$?

12. ## Re: Algebra and Functions

Originally Posted by jacobjacob44
Ahh I see! So would that make the final answer be $\frac{9}{6}$?
Is 2^2 = 6?

13. ## Re: Algebra and Functions

apologies, $\frac{9}{4}$

14. ## Re: Algebra and Functions

Originally Posted by jacobjacob44
apologies, $\frac{9}{4}$
you got it!

15. ## Re: Algebra and Functions

Thanks for all of the help.