# Exponents

• Mar 2nd 2011, 04:42 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, 04:56 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?

$12=2^2\cdot 3$ so what is $(2^2\cdot 3)^5~?$
• Mar 2nd 2011, 05:03 AM
dumluck
Quote:

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

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, 05:32 AM
angypangy
$27^\frac{1}{3} = \sqrt[3]{27} and you know what 27^4 means. Just split it into parts. It doesn't matter whether you evaluate the fractional part first or the part first$
• Mar 2nd 2011, 05:37 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?

$(27)^{\frac{4}{3}}=(3^3)^{\frac{4}{3}}=3^4$
• Mar 2nd 2011, 05:45 AM
dumluck
Quote:

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

Thanks Plato so $(27)^{\frac{4}{3}}$ 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, 05:47 AM
Wilmer
27^4 / 3 or 27^(4/3) ?
• Mar 2nd 2011, 05:51 AM
dumluck
Quote:

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

Neither.. (3SQRT(27))^4
• Mar 2nd 2011, 06:49 AM
Plato
Quote:

Originally Posted by dumluck
$(27)^{\frac{4}{3}}$ 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?

$(N^x)^y=N^{xy}$
$\left( {\left[ {\sqrt[3]{{x^2 }}} \right]} \right)^4 = x^{\frac{8}
{3}} = x^2 \sqrt[3]{{x^2 }}$
• Mar 2nd 2011, 07:14 AM
dumluck
Quote:

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

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

IS that correct?
• Mar 2nd 2011, 07:18 AM
Plato
Quote:

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

YES.
But also $3\left(\frac{1}{3}\right)=1$
• Mar 2nd 2011, 09:27 AM
dumluck
Quote:

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

Thanks plato, where does that come in?
• Mar 2nd 2011, 11:14 AM
Plato
Quote:

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

$\sqrt[3]{{x^3 }} = \left( {x^3 } \right)^{\frac{1}{3}}$
• Mar 3rd 2011, 02:56 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 " $27^{4/3}$" 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, 04:57 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 " $27^{4/3}$" 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.