# Evaluating Fraction/exponent...

• March 25th 2011, 02:05 PM
Freaked
Evaluating Fraction/exponent...
This is really driving me nuts, I've spent about 2 hours on and off trying to get to the given answer...With no close results.
Apologies for lack of formatted maths symbols.

(8/27)^3/2

The answer is given as (16/243)√6

I follow the normal process, as in: = √(8/27)^3
then go down the ((√2.√4)/(√3.√9))^3
= ((2√2)/(3√3))^3

This takes me nowhere near to (16/243)√6

If somone has the time, could you please show me how I can get to the (16/243)√6 and maybe explain the error of my process too.

It would be much appreciated

Thank you
• March 25th 2011, 02:13 PM
mathfun
${\left( {\tfrac{8}{{27}}} \right)^{\tfrac{3}{2}}} = {\left( {\tfrac{{{2^3}}}{{{3^3}}}} \right)^{\tfrac{3}{2}}} = {\left( {\tfrac{2}{3}} \right)^{\tfrac{9}{2}}} = {\left( {\tfrac{{\sqrt 6 }}{3}} \right)^9} = \left( {\tfrac{{\sqrt 6 }}{3}} \right){\left( {\tfrac{{\sqrt 6 }}{3}} \right)^8} = \left( {\tfrac{{\sqrt 6 }}{3}} \right) \cdot \tfrac{{{2^4} \cdot {3^4}}}{{{3^8}}} = \sqrt 6 \cdot \tfrac{{{2^4}}}{{{3^9}}} = \sqrt 6 \tfrac{{16}}{{243}}$
• March 25th 2011, 02:28 PM
Freaked
ok, thanks : )

Why is it that you would have (√6/3)^9 and not (√6/√3)^9.
Maybe I just can't see the wood for the trees now, after so long tacklilng this one... : /

Edit:
well, infact, I can't see how you get to the (√6/3)^9 step, I can't see where the √6 comes from looking at the previous step...Shouldn't (2/3)^9/2 = √(2/3)^9 ? ...My poor head
• March 25th 2011, 02:47 PM
skeeter
$\left(\dfrac{8}{27}\right)^{\frac{3}{2}} = \left(\dfrac{2^3}{3^3}\right)^{\frac{3}{2}} = \left[\left(\dfrac{2}{3}\right)^3\right]^{\frac{3}{2}} = \left(\dfrac{2}{3}\right)^{\frac{9}{2}} = \dfrac{(\sqrt{2})^9}{(\sqrt{3})^9} = \dfrac{16\sqrt{2}}{81\sqrt{3}} = \dfrac{16\sqrt{6}}{243}$
• March 25th 2011, 02:59 PM
Freaked
Ok, this seems more like my line of thinking - until I get to 16√2/81√3. Is it possible to show why (√2)^9/(√3)^9 can = 16√2/81√3 ? I'm sorry if I'm just being slow...
• March 25th 2011, 03:06 PM
e^(i*pi)
$(\sqrt{2})^9 = \sqrt{2} \cdot (\sqrt{2})^8 = \sqrt{2} \cdot 2^{8/2} = 2^4 \cdot \sqrt{2} = 16\sqrt{2}$

Same principle applies for the demoninator
• March 25th 2011, 03:15 PM
Freaked
I finally can see the logic.
This is why Maths is so appealing! I just was unable to follow this line of reasoning beforehand.

Thank you kindly !