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Math Help - Solve for x

  1. #1
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    Solve for x

    how do you solve

    3^(2x)-3^(x+2)+8=0
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  2. #2
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    i think this needs a change of base on the 8

    8 = 3^log.3. 8

    in general

    a^x = b^(log.b. a)(x)


    move the term with the negative to the other side so they are all positive
    then since you have all the numbers with base 3 you can say the bases are equal therefore the exponents must be equal

    and then you can equate the exponents
    i get x = 2 - log.3. 8
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by fatman View Post
    how do you solve

    3^(2x)-3^(x+2)+8=0
    lets rewrite this as

    (3^x)^2-3^23^x+8=0

    this is quadratic in 3^x so to simplify let u=3^x

    The we get

    u^2-9u+8=0 \iff (u-1)(u-8)=0 so u =1 and u=8

    but u=3^x so we get

    1=3^x \iff \log_3(1)= \log_33^x \iff 0=x

    8=3^x \iff \ln(8)=\ln(3^x) \iff \ln(8)=x\ln(3) \iff \frac{\ln(8)}{\ln(3)}=x

    Good luck.
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  4. #4
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    oh i think my mistake was when i said that since the exponents of
    A and B and C are all the same in the relation:

    A+B=C

    when in fact it only works when you have a relation:
    A=B

    anyone confirm?
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  5. #5
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    my mistake, i made a blunder with the exponent rules
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  6. #6
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    Thanks

    Thank you very much for your all your help with quick replies
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