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Thread: Backward Exponets

  1. #1
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    Backward Exponets



    i understood a,b and c but d made not alot of sense. how do i go backwards? thanks
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  2. #2
    Super Member Quacky's Avatar
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    Hint:
    $\displaystyle \sqrt x = x^\frac{1}{2}$
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  3. #3
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    still confused.
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  4. #4
    Super Member Quacky's Avatar
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    $\displaystyle \sqrt{64}=8$.
    So, using what I've said above, to what power must you raise 64 to get 8?
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  5. #5
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    64 ^ .5 = 8

    thanks


    i cant do 64 ^ .5 to get 2 so what do I do?
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  6. #6
    Super Member Quacky's Avatar
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    It's better to think about the exponents as fractions.

    If $\displaystyle 2^2=4$
    Then $\displaystyle 4^\frac{1}{2}=2$

    If $\displaystyle 9^4=6561$

    Then $\displaystyle 6561^\frac{1}{4}=9$

    So it's easiest to work backwards.

    $\displaystyle 3^4=81$
    So $\displaystyle 81^?=3$
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  7. #7
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    For G, H and I remember that an exponent raised to an exponent is just the two exponents multiplied together.

    so,

    x^2^6=x^12
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  8. #8
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    so to get 81 to 3 you do 81^.25

    hjowever for 64 -> 2 how do i do 2^1/6 to check?
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  9. #9
    Super Member Quacky's Avatar
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    That's correct!
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  10. #10
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    Quote Originally Posted by Quacky View Post
    That's correct!
    awsome, thank you.



    i know its unrelated, but how do i do these?

    for a i have

    .5 = x

    is that correct? thanks
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  11. #11
    Super Member Quacky's Avatar
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    $\displaystyle 125^x=25^{x+1}$

    I would do that by writing both as 5^{something}
    $\displaystyle 125=5^3$
    $\displaystyle 25=5^2$

    So $\displaystyle (5^3)^x=(5^2)^{x+1}$
    $\displaystyle 5^{3x}=5^{2x+2}$
    So:
    $\displaystyle 3x=2x+2$
    $\displaystyle x=2$

    Part b) is harder. Try to write it out as powers of 3.
    For part c) Try to write it out as powers of 2
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  12. #12
    Member integral's Avatar
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    Is this not a logarithm problem?
    $\displaystyle 3^x=27 \therefore log_3 27=x$
    then

    $\displaystyle x=\frac{log27}{log3}$
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  13. #13
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    Quote Originally Posted by minneola24 View Post
    awsome, thank you.

    -snip-

    i know its unrelated, but how do i do these?

    for a i have

    .5 = x

    is that correct? thanks
    A is x=2. 125 is 25^2, so 25^3 = 125^2

    As for B, I have the answer but I can't explain why it works so I won't post it.
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  14. #14
    Super Member Bacterius's Avatar
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    Quote Originally Posted by integral View Post
    Is this not a logarithm problem?
    $\displaystyle 3^x=27 \therefore log_3 27=x$
    then

    $\displaystyle x=\frac{log27}{log3}$
    Don't confuse the OP, integral (even your signature seems to be an extension of your answer xD)

    For part c), you could note that $\displaystyle 4 = 2 \times 2$, and try to find yourself with just two's on the right side. Then note that $\displaystyle 8 = 2^3$. The rest should follow.
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