anonymous
  • anonymous
@jdoe0001 How do you solve for the equation log x+ log(x-3) = 1?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
@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.
jdoe0001
  • jdoe0001
hmmm
anonymous
  • anonymous
@jdoe0001 actually I just graphed it on my graphing calculator. x=5

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jdoe0001
  • jdoe0001
right... I get something else one sec
anonymous
  • anonymous
jdoe0001
  • jdoe0001
\(log(x)+log(x-3)=1\implies log[(x)(x-3)]=1 \\ \quad \\ log_{{\color{brown}{ 10}}}[(x)(x-3)]=1\implies {\color{brown}{ 10}}^1=(x)(x-3) \\ \quad \\ 10=x^2-3x\implies 0=x^2-3x-10 \\ \quad \\ 0=(x-5)(x+2)\implies \begin{cases} x=5\\ x=-2 \end{cases}\)
anonymous
  • anonymous
@jdoe0001 are you sure about x=-2? I plugged -2 in for x and I got "Math ERROR" in the calculator
jdoe0001
  • jdoe0001
right... logarithms do not take a negative value, is 0 or up so ends up being 5 :)
jdoe0001
  • jdoe0001
but what the answer sheet shows.. looks very different

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