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Math Help - Exponents

  1. #1
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    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?

    Thanks In Advance,
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  2. #2
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    Quote Originally Posted by dumluck View Post
    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~?
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  3. #3
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    Quote Originally Posted by Plato View Post
    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.
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  4. #4
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     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
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  5. #5
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    Quote Originally Posted by dumluck View Post
    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
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  6. #6
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    Quote Originally Posted by Plato View Post
    (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
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  7. #7
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    27^4 / 3 or 27^(4/3) ?
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  8. #8
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    Quote Originally Posted by Wilmer View Post
    27^4 / 3 or 27^(4/3) ?
    Neither.. (3SQRT(27))^4
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  9. #9
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    Quote Originally Posted by dumluck View Post
    (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 }}
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  10. #10
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    Quote Originally Posted by Plato View Post
    (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?
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  11. #11
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    Quote Originally Posted by dumluck View Post
    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
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  12. #12
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    Quote Originally Posted by Plato View Post
    YES.
    But also 3\left(\frac{1}{3}\right)=1
    Thanks plato, where does that come in?
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  13. #13
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    Quote Originally Posted by dumluck View Post
    Thanks plato, where does that come in?
    \sqrt[3]{{x^3 }} = \left( {x^3 } \right)^{\frac{1}{3}}
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  14. #14
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    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!)
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  15. #15
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    Quote Originally Posted by HallsofIvy View Post
    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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