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Math Help - logarithm problem

  1. #1
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    logarithm problem

    logarithm problem
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    logarithm problem
    The first question can be transformed into a quadratic in 2^x+2^{-x} , by observing that:

    (2^x+2^{-x})^2=2^{2x}+2+2^{-2x}=[4^x+4^{-x}]+2

    So the equation becomes:

     <br />
2[2^x+2^{-x}]^2 -7[2^x+2^{-x}]=8=0<br />

    Now solve this quadratic in 2^x+2^{-x} and proceed from there.

    RonL
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  3. #3
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    Quote Originally Posted by afeasfaerw23231233 View Post
    logarithm problem
    Hello,

    (2^x+2^{-x})^2=2^{2x}+2+2^{-2x} = (4^x+4^{-x})+2

    Therefore use (2^x+2^{-x}) = y. Then (4^x+4^{-x}) = y^2-2 . Your equation becomes:

    2(y^2-2)-7y+10=0 . Now continue. Don't forget to re-substitute to calculate x.

    = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

    With the second problem you can use the property of logarithms:

    \log_b(a)=c~\implies~\log_a(b)=\frac1c~\implies~\f  rac1{\log_a(b)} = c
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  4. #4
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    Hello, afeasfaerw23231233!

    \log_x4 - \log_2x \:=\:1
    From the Base-Change Formula: . \log_x4 \:=\:\frac{\log_24}{\log_2x}\:=\:\frac{2}{\log_2x}

    The equation becomes: . \frac{2}{\log_2x} - \log_2x \:=\:1

    . . which simplifies to: . \left(\log_2x\right)^2 + \log_2x - 2 \:=\:0

    . . which factors: . \left(\log_2x + 2\right)\left(\log_2x - 1\right) \:=\:0


    Then: . \log_2x + 2 \:=\:0\quad\Rightarrow\quad \log_2x \:=\:-2 . . . no real roots

    .And: . \log_2x - 1\:=\:0\quad\Rightarrow\quad \log_2x \:=\:1\quad\Rightarrow\quad \boxed{x \:=\:2}

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