Exponents

**dumluck** · March 2nd 2011, 03:42 AM

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?

Thanks In Advance,

**Plato** · March 2nd 2011, 03:56 AM

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 $(2^2\cdot 3)^5~?$

**dumluck** · March 2nd 2011, 04:03 AM

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.

**angypangy** · March 2nd 2011, 04:32 AM

$27^\frac{1}{3} = \sqrt[3]{27} and you know what 27^4 means. Just split it into parts. It doesn't matter whether you evaluate the fractional part first or the part first$

**Plato** · March 2nd 2011, 04:37 AM

Originally Posted by dumluck

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

$(27)^{\frac{4}{3}}=(3^3)^{\frac{4}{3}}=3^4$

**dumluck** · March 2nd 2011, 04:45 AM

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

**Wilmer** · March 2nd 2011, 04:47 AM

27^4 / 3 or 27^(4/3) ?

**dumluck** · March 2nd 2011, 04:51 AM

Originally Posted by Wilmer

27^4 / 3 or 27^(4/3) ?

Neither.. (3SQRT(27))^4

**Plato** · March 2nd 2011, 05:49 AM

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?

$(N^x)^y=N^{xy}$
$\left( {\left[ {\sqrt[3]{{x^2 }}} \right]} \right)^4 = x^{\frac{8}<br /> {3}} = x^2 \sqrt[3]{{x^2 }}$

**dumluck** · March 2nd 2011, 06:14 AM

Originally Posted by Plato

$(N^x)^y=N^{xy}$
$\left( {\left[ {\sqrt[3]{{x^2 }}} \right]} \right)^4 = x^{\frac{8}<br /> {3}} = x^2 \sqrt[3]{{x^2 }}$

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

IS that correct?

**Plato** · March 2nd 2011, 06:18 AM

Originally Posted by dumluck

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

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

**dumluck** · March 2nd 2011, 08:27 AM

Originally Posted by Plato

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

Thanks plato, where does that come in?

**Plato** · March 2nd 2011, 10:14 AM

Originally Posted by dumluck

Thanks plato, where does that come in?

$\sqrt[3]{{x^3 }} = \left( {x^3 } \right)^{\frac{1}{3}}$

**HallsofIvy** · March 3rd 2011, 01:56 AM

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!)

**dumluck** · March 3rd 2011, 03:57 AM

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.

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