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anonymous

  • 3 years ago

solve a log

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  1. anonymous
    • 3 years ago
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    \[\log_{a}x+\log_{a}x-2=\log_{a}x+4 \]\ \[\log_{a}x(x-2)=\log_{a}x+4\] that is my work so far I do not know how to convert to exponential

  2. anonymous
    • 3 years ago
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    I am wondering if I divide by \[\log_{a}x+4\] and then have it equal to zero which means my final answer is just 1

  3. anonymous
    • 3 years ago
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    assuming the right hand side is \(\log_a(x+4)\) then your last job is to solve \[x(x-2)=x+4\]

  4. anonymous
    • 3 years ago
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    so the logs just drop out

  5. anonymous
    • 3 years ago
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    log is a "one to one function" so if \[\log(A)=\log(B)\] then \(A=B\)

  6. anonymous
    • 3 years ago
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    ok I guess but I dont really understand

  7. anonymous
    • 3 years ago
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    ok suppose instead of log, you had \[f(x)=2x+4\]

  8. anonymous
    • 3 years ago
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    if you were told that \[2a+4=2b+4\] then you would know that \(a=b\) right?

  9. anonymous
    • 3 years ago
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    it is the same with the log if \(\log(x)=\log(y)\) then you know \(x=y\)

  10. anonymous
    • 3 years ago
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    in other words, as in your example, the only way for \(\log(x^2-2x)\) to be equal to \(\log(x+4)\) is if \(x^2-2x=x+4\)

  11. anonymous
    • 3 years ago
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    oh I see now so if i understand correctly it is because we know they must be equal that the fact that it is a log of each does not matter because no matter what they yield the same result

  12. anonymous
    • 3 years ago
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    yes, exactly

  13. anonymous
    • 3 years ago
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    Thank you I appreciate your help

  14. anonymous
    • 3 years ago
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    if i know \(\log(x)=\log(5)\) then i know that \(x=5\) i cannot be anything else

  15. anonymous
    • 3 years ago
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    yw

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