1. ## 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?

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

3. 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.

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

5. Originally Posted by dumluck
27^4/3 = 3SQRT(27^4) = (3SQRT(27))^4 ... how do we get 3^4 from the last?
$(27)^{\frac{4}{3}}=(3^3)^{\frac{4}{3}}=3^4$

6. 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

7. 27^4 / 3 or 27^(4/3) ?

8. Originally Posted by Wilmer
27^4 / 3 or 27^(4/3) ?
Neither.. (3SQRT(27))^4

9. 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?
$(N^x)^y=N^{xy}$
$\left( {\left[ {\sqrt[3]{{x^2 }}} \right]} \right)^4 = x^{\frac{8}
{3}} = x^2 \sqrt[3]{{x^2 }}$

10. 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?

11. 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 $3\left(\frac{1}{3}\right)=1$

12. Originally Posted by Plato
YES.
But also $3\left(\frac{1}{3}\right)=1$
Thanks plato, where does that come in?

13. Originally Posted by dumluck
Thanks plato, where does that come in?
$\sqrt[3]{{x^3 }} = \left( {x^3 } \right)^{\frac{1}{3}}$

14. 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!)

15. 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.