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Thread: 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 $\displaystyle 2^x+2^{-x} $, by observing that:

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

    So the equation becomes:

    $\displaystyle
    2[2^x+2^{-x}]^2 -7[2^x+2^{-x}]=8=0
    $

    Now solve this quadratic in $\displaystyle 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,

    $\displaystyle (2^x+2^{-x})^2=2^{2x}+2+2^{-2x} = (4^x+4^{-x})+2$

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

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

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

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

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

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

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

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

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


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

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

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