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anonymous

  • 5 years ago

can someone check this out?

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  1. anonymous
    • 5 years ago
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    What?

  2. anonymous
    • 5 years ago
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    \[2\log_{2}x-\log_{2}(x+3)=2\]

  3. anonymous
    • 5 years ago
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    i got \[x=2/\log_{2}(x+3) \] i just need to see if its correct?

  4. anonymous
    • 5 years ago
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    noit is not correct

  5. anonymous
    • 5 years ago
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    means solve for x, yes?

  6. anonymous
    • 5 years ago
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    yes.

  7. anonymous
    • 5 years ago
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    ok first combine the left hand side into one logarithm using things like\[log(A) + log(B) = log(AB)\]

  8. anonymous
    • 5 years ago
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    first \[2log_2(x)=log_2(x^2)\]

  9. anonymous
    • 5 years ago
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    then \[log_2(x^2)-log_2(x+3)=log_2(\frac{x^2}{x+3})\]

  10. anonymous
    • 5 years ago
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    so we have \[log_2(\frac{x^2}{x+3})=2\] in equivalent exponential form this means \[\frac{x^2}{x+3}=2^2=4\]

  11. anonymous
    • 5 years ago
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    \[x^2=4(x+3)\] \[x^2=4x+12\] \[x^2-4x-12=0\]

  12. anonymous
    • 5 years ago
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    \[(x-6)(x+2)=0\] \[x=6\]or \[x=-2\] but we throw out the -2 because you cannot take the log of a negative number. so answer is \[x=6\]

  13. anonymous
    • 5 years ago
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    hope the steps are clear.

  14. anonymous
    • 5 years ago
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    thank you..

  15. anonymous
    • 5 years ago
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    welcome!

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