simplify radicals and fractional exponents?

Question

This is the equation, please help me. I don't have any idea what to do with this. Thank you.

√((32)^((4)/(5))(2^(n+1))^(4))/8^3 (16)^(n-1)

That is a whole fraction within a radical sign all over √6:(9)

$$\frac{\sqrt{\frac{32^{\frac45}(2^{(n+1)})^4}{8^316^{(n-1)}}}}{\sqrt[6] 9}$$

Did you mean this : $$\sqrt{((32)^{\frac45}(2^{n+1})^4)/8^3 (16)^{n-1} }$$? — lab bhattacharjee, Aug 18 '13 at 11:43
I have put a version of your radical underneath what you wrote in the hope of clarifying what you are trying to compute. Could you please let me know in a comment whether this is correct, or what is wrong with it. — Mark Bennet, Aug 18 '13 at 11:45
Is your expression $\sqrt{\dfrac{32^{4/5}\cdot(2^{n+1})^4}{8^3 \cdot 16^{n-1}}}$ divided by $\sqrt{\dfrac69}$? — Alraxite, Aug 18 '13 at 11:45
put down the other equation, put it like a denominator and after which, make sixth root of 9 as a denominator again — ikaidubidu, Aug 18 '13 at 11:46
alraxite, it is a sixthe root of 9 and without the radical sign — ikaidubidu, Aug 18 '13 at 11:47
@labbhattacharjee - thanks for the comment. Note - what I have put needs an (n-1) as the power of 16 - but since there are competing edits and it is taking time for things to load on my computer I'll leave that to be changed in due course. — Mark Bennet, Aug 18 '13 at 11:48
So your expression is $\sqrt{\dfrac{32^{4/5}\cdot(2^{n+1})^4}{8^3 \cdot 16^{n-1}}}$ divided by$\sqrt[9]6$? — Alraxite, Aug 18 '13 at 11:50
So this: $\sqrt{\dfrac{32^{4/5}\cdot(2^{n+1})^4}{8^3 \cdot 16^{n-1}}}/(\sqrt[6]{9})$? — Alraxite, Aug 18 '13 at 11:52
oh wait, the 16 should have the parenthesis, not the exponent :) @MarkBennet — ikaidubidu, Aug 18 '13 at 12:06
@ikaidubidu I think the parentheses here are not wholly necessary except for clarifying the order of operations. But the way the expression is now written would do that without the parentheses. — Mark Bennet, Aug 18 '13 at 12:26

Alraxite · Answer 1 · 2013-08-18T12:24:43.410

1

First let's simplify the numerator:

$\sqrt{\dfrac{32^{4/5}\cdot(2^{n+1})^4}{8^3 \cdot 16^{n-1}}} =\sqrt{\dfrac{(2^5)^{4/5}\cdot(2^{n+1})^4}{(2^3)^3 \cdot (2^4)^{n-1}}} =\sqrt{\dfrac{2^4\cdot 2^{4n+4}}{2^9 \cdot 2^{4n-4}}} =\sqrt{2^{4-9+(4n+4)-(4n-4)}}=\sqrt{2^3}=2^{\frac32}$.

The denominator:

$\sqrt[6]9=9^{\frac16}=(3^2)^{\frac16}=3^{\frac13}$.

So, your expression is $\dfrac{2^\frac32}{3^\frac13}$. You can rewrite this as $\dfrac{2^\frac96}{3^\frac26}=\sqrt[6]{\dfrac{2^9}{3^2}}=\sqrt[6]{\dfrac{512}{9}}$.

edited Aug 18 '13 at 12:24

answered Aug 18 '13 at 12:05

Alraxite

5,647

wait, on the numerator u had it 3/2 then u wrote 2/3 on the final expression? – ikaidubidu Aug 18 '13 at 12:18
@ikaidubidu Oh, you are correct! – Alraxite Aug 18 '13 at 12:19
is it a 3/2 or a 2/3? – ikaidubidu Aug 18 '13 at 12:20
@ikaidubidu I've edited my answer. It's a $\frac32$. – Alraxite Aug 18 '13 at 12:26
can i ask how did you come up with an exponent of 3 on the 4-9...etc etc ? – ikaidubidu Aug 18 '13 at 12:34
@ikaidubidu The expression $4-9+(4n+4)-(4n-4)$ simplifies to $3$. Because $4-9+(4n+4)-(4n-4)= -5 + 4n + 4 -4n +4=-5+8=3$. – Alraxite Aug 18 '13 at 12:36
thank you @Alraxite, now im able to discuss it tomorrow :) – ikaidubidu Aug 18 '13 at 12:47

simplify radicals and fractional exponents?

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