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Math Help - Logarithem equation

  1. #1
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    Logarithem equation

    please help me with this equation:



    Find the value of X such that:

    log2x + log232 = 6logx2


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  2. #2
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    <br />
\log _2 x + \log _2 32 = 6\log _x 2 \hfill \\

     \Rightarrow \log _2 \left( {x \times 32} \right) = 6\left( {\frac{{\log 2}}<br />
{{\log x}}} \right) \hfill

    \Rightarrow \log _2 \left( {32x} \right) = \frac{{6\log 2}}<br />
{{\log x}} \hfill \\

    \Rightarrow \log _2 \left( {32x} \right) = \frac{{\log \left( 2 \right)^6 }}<br />
{{\log x}} \hfill \\

    \Rightarrow \frac{{\log \left( {32x} \right)}}<br />
{{\log 2}} = \frac{{\log 64}}<br />
{{\log x}} \hfill \\

    \Rightarrow x = 2
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  3. #3
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    thx, but I did not understand how u got x=2 from
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  4. #4
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    Hello,
    Quote Originally Posted by Musab View Post
    thx, but I did not understand how u got x=2 from
    By observation.

    Or another method :
    \log_2(x)+\log_2(32)=6 \log_x(2)

    32=2^5 ---> \log_2(32)=5

    \frac{\ln(x)}{\ln(2)}+5=6 \frac{\ln(2)}{\ln(x)}

    Multiply the two sides by \ln(2) \cdot \ln(x) and solve the quadratic
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  5. #5
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    Hello, Musab!

    Find the value of x such that: . \log_2(x) + \log_2(32) \:=\:6\log_x(2)
    Recall the theorem: . \log_a(b)) \:=\:\frac{1}{\log_b(a)}

    We have: . \log_2(x) + 5 \;=\;\frac{6}{\log_2(x)}


    Then: . \left[\log_2(x)\right]^2 + 5\left[\log_2(x)\right] \:=\:6 \quad\Rightarrow\quad [\log_2(x)]^2 + 5\left[\log_2(x)\right] - 6 \:=\:0


    Factor: . \left[\log_2(x)+6\right]\,\left[\log_2(x) - 1\right] \:=\:0


    Then: . \log_2(x) +6 \:=\:\quad\Rightarrow\quad \log_2(x) \:=\:-6 \quad\Rightarrow\quad x \:=\:2^{-6} \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{64}}

    And: . \log_2(x) -1 \:=\: \quad\Rightarrow\quad \log_2(x) \:=\:1\quad\Rightarrow\quad x \:=\:2^1 \quad\Rightarrow\quad \boxed{x\:=\:2}



    Edit: corrected an awful blunder . . .
    .
    Last edited by Soroban; September 4th 2008 at 01:39 PM.
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