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anonymous

  • one year ago

@jdoe0001 How do you solve for the equation log x+ log(x-3) = 1?

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  1. anonymous
    • one year ago
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    @jdoe0001 I know the answer is x= .5(3+sqrt of (9+4e)). I just don't know how the answer key got that answer.

  2. jdoe0001
    • one year ago
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    hmmm

  3. anonymous
    • one year ago
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    @jdoe0001 actually I just graphed it on my graphing calculator. x=5

  4. jdoe0001
    • one year ago
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    right... I get something else one sec

  5. anonymous
    • one year ago
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    ok @jdoe0001

  6. jdoe0001
    • one year ago
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    \(log(x)+log(x-3)=1\implies log[(x)(x-3)]=1 \\ \quad \\ log_{{\color{brown}{ 10}}}[(x)(x-3)]=1\implies {\color{brown}{ 10}}^1=(x)(x-3) \\ \quad \\ 10=x^2-3x\implies 0=x^2-3x-10 \\ \quad \\ 0=(x-5)(x+2)\implies \begin{cases} x=5\\ x=-2 \end{cases}\)

  7. anonymous
    • one year ago
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    @jdoe0001 are you sure about x=-2? I plugged -2 in for x and I got "Math ERROR" in the calculator

  8. jdoe0001
    • one year ago
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    right... logarithms do not take a negative value, is 0 or up so ends up being 5 :)

  9. jdoe0001
    • one year ago
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    but what the answer sheet shows.. looks very different

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spraguer (Moderator)
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