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Math Help - Divinding exponents

  1. #1
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    Divinding exponents

    Hey guys, for the following question how do I solve for x?

    (2^3)^x = 2^6

    I know that to raise a power to a power I need to multiply the exponents. In the above question I thought that maybe I can solve it by dividing the exponents, but my math book doesn't mention anything about dividing exponents.

    here's how I tried to solve it:
    (2^3)^x = 2^6
    (2^3)^x = (2^3)^2
    x = 2

    I assumed that since both have the same bases, I can equate x to 2.
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  2. #2
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    Re: Divinding exponents

    Your solution is correct.

    I assumed that since both have the same bases, I can equate x to 2.
    Yes, exponentiation is a one-to-one function: if a > 1 or 0 < a < 1, then a^x = a^y implies x = y for all x and y.
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  3. #3
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    Re: Divinding exponents

    Thanks emakarov. I have another question: is it possible to divide exponents?
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  4. #4
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    Re: Divinding exponents

    Quote Originally Posted by darksoulzero View Post
    I have another question: is it possible to divide exponents?
    What do you mean by this? The right question in mathematics is whether something is true, not if one can do something.
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  5. #5
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    Re: Divinding exponents

    Quote Originally Posted by darksoulzero View Post
    I have another question: is it possible to divide exponents?
    If you mean a^x / a^y, then: a^(x-y)
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  6. #6
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    Re: Divinding exponents

    Ok, I see thanks guys.

    based on what emakarov said, if a>1, or 0<a<1, a^x = a^y, then x = y, is the following also true?

    x^a = y^a, then x = y?

    here's what i tried

    \\x^3 = 0.216\\x^3 = \sqrt[3]{0.216} \\x^3 = (0.6)^3\\x = 0.6
    Last edited by darksoulzero; July 16th 2011 at 01:22 PM.
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  7. #7
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    Re: Divinding exponents

    And also: a^x=a^y \Rightarrow x=y.

    Your solution is right, but your second step is incorrect, because if you say: x^3=\sqrt[3]{0,216} that means x^3=0,6.
    It has to be.
    x^3=0,216
    \Leftrightarrow x=\sqrt[3]{0,216}
    \Leftrightarrow x=0,6.
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  8. #8
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    Re: Divinding exponents

    What I meant for the second step was this:

    \\x^3 = 0.216\\\sqrt[3]{x^3} = \sqrt[3]{0.216}\\ x = 0.6

    And like I said, is teh following true?

    x^a = 3^a, x = 3?
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  9. #9
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    Re: Divinding exponents

    Yes, you were told that in the very first response:

    emakarov said
    "Yes, exponentiation is a one-to-one function: if a > 1 or 0 < a < 1, then a^x = a^y implies x = y for all x and y. "
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  10. #10
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    Re: Divinding exponents

    Quote Originally Posted by darksoulzero View Post
    based on what emakarov said, if a>1, or 0<a<1, a^x = a^y, then x = y, is the following also true?

    x^a = y^a, then x = y?
    Yes, if a\ne 0 and x, y > 0. In fact, this follows from the previous statement. Indeed, x^a=e^{a\ln x}, so if x^a=y^a, then e^{a\ln x}=e^{a\ln y}, so a\ln x=a\ln y and thus \ln x=\ln y if a\ne 0. Now, ln is also a one-to-one function, so x = y.
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  11. #11
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    Re: Divinding exponents

    Ok, I see. thanks emakarov. And thanks to everyone else that helped.
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