# Exponents

• Mar 2nd 2011, 03:42 AM
dumluck
Exponents
Hi All,
Quite a Simply question but elluding me currently.

I know how 2^4.3^4 = 6^4

But how does 12^5 = 2^10.3^5?

I understand it could also be 2^5.2^6 but I'm trying to determine the above answer?

• Mar 2nd 2011, 03:56 AM
Plato
Quote:

Originally Posted by dumluck
I know how 2^4.3^4 = 6^4
But how does 12^5 = 2^10.3^5?

$\displaystyle 12=2^2\cdot 3$ so what is $\displaystyle (2^2\cdot 3)^5~?$
• Mar 2nd 2011, 04:03 AM
dumluck
Quote:

Originally Posted by Plato
$\displaystyle 12=2^2\cdot 3$ so what is $\displaystyle (2^2\cdot 3)^5~?$

ah yes so it's (4.3)^5 which is (2^2.3)5 , which is 2^10.3^5.

Thank you. Can I ask one more

27^4/3 = 3SQRT(27^4) = (3SQRT(27))^4 ... how do we get 3^4 from the last?

Thanks so much.
• Mar 2nd 2011, 04:32 AM
angypangy
$\displaystyle 27^\frac{1}{3} = \sqrt[3]{27} and you know what 27^4 means. Just split it into parts. It doesn't matter whether you evaluate the fractional part first or the part first$
• Mar 2nd 2011, 04:37 AM
Plato
Quote:

Originally Posted by dumluck
27^4/3 = 3SQRT(27^4) = (3SQRT(27))^4 ... how do we get 3^4 from the last?

$\displaystyle (27)^{\frac{4}{3}}=(3^3)^{\frac{4}{3}}=3^4$
• Mar 2nd 2011, 04:45 AM
dumluck
Quote:

Originally Posted by Plato
$\displaystyle (27)^{\frac{4}{3}}=(3^3)^{\frac{4}{3}}=3^4$

Thanks Plato so $\displaystyle (27)^{\frac{4}{3}}$ is the same as 3SQRT(27)^4? are we then just 'crossing out' the 3's? if so; why? I.e. what exponent rule govens it?

Thanks
• Mar 2nd 2011, 04:47 AM
Wilmer
27^4 / 3 or 27^(4/3) ?
• Mar 2nd 2011, 04:51 AM
dumluck
Quote:

Originally Posted by Wilmer
27^4 / 3 or 27^(4/3) ?

Neither.. (3SQRT(27))^4
• Mar 2nd 2011, 05:49 AM
Plato
Quote:

Originally Posted by dumluck
$\displaystyle (27)^{\frac{4}{3}}$ is the same as 3SQRT(27)^4? are we then just 'crossing out' the 3's? if so; why? I.e. what exponent rule govens it?

$\displaystyle (N^x)^y=N^{xy}$
$\displaystyle \left( {\left[ {\sqrt[3]{{x^2 }}} \right]} \right)^4 = x^{\frac{8} {3}} = x^2 \sqrt[3]{{x^2 }}$
• Mar 2nd 2011, 06:14 AM
dumluck
Quote:

Originally Posted by Plato
$\displaystyle (N^x)^y=N^{xy}$
$\displaystyle \left( {\left[ {\sqrt[3]{{x^2 }}} \right]} \right)^4 = x^{\frac{8} {3}} = x^2 \sqrt[3]{{x^2 }}$

Thanks Plato so using this.
$\displaystyle \left( {\left[ {\sqrt[3]{{27}}} \right]} \right)^4 = {\left[ {\sqrt[3]{{3^3}}} \right]} \right)^4 = 3^{\frac{12} {3}} ?$ and diving 12 by3 we get $\displaystyle 3^4$

IS that correct?
• Mar 2nd 2011, 06:18 AM
Plato
Quote:

Originally Posted by dumluck
Thanks Plato so using this.
$\displaystyle \left( {\left[ {\sqrt[3]{{27}}} \right]} \right)^4 = {\left[ {\sqrt[3]{{3^3}}} \right]} \right)^4 = 3^{\frac{12} {3}} ?$ and diving 12 by3 we get $\displaystyle 3^4$
IS that correct?

YES.
But also $\displaystyle 3\left(\frac{1}{3}\right)=1$
• Mar 2nd 2011, 08:27 AM
dumluck
Quote:

Originally Posted by Plato
YES.
But also $\displaystyle 3\left(\frac{1}{3}\right)=1$

Thanks plato, where does that come in?
• Mar 2nd 2011, 10:14 AM
Plato
Quote:

Originally Posted by dumluck
Thanks plato, where does that come in?

$\displaystyle \sqrt[3]{{x^3 }} = \left( {x^3 } \right)^{\frac{1}{3}}$
• Mar 3rd 2011, 01:56 AM
HallsofIvy
Please do not write "3sqrt" for the cube root! The "sq" in "sqrt" is from square and the cube root has nothing to do with a square root! For you to say you did NOT mean "$\displaystyle 27^{4/3}$" but meant "(3sqrt(27))^4" is exceptionally confusing.

(Yes, I know that the LaTeX for "cube root" is "\sqrt[3]{}" and I am just as annoyed at that!)
• Mar 3rd 2011, 03:57 AM
dumluck
Quote:

Originally Posted by HallsofIvy
Please do not write "3sqrt" for the cube root! The "sq" in "sqrt" is from square and the cube root has nothing to do with a square root! For you to say you did NOT mean "$\displaystyle 27^{4/3}$" but meant "(3sqrt(27))^4" is exceptionally confusing.

(Yes, I know that the LaTeX for "cube root" is "\sqrt[3]{}" and I am just as annoyed at that!)

No problem, will look up LaTex to ensure I'm describing correctly.