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GOODMAN

  • 3 years ago

Have a few problems I was stuck on. 1)log_9(x+4)+log_9(x-4)=1 2)Sqrt(3x-2)-sqrt(x)=1

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  1. GregTheo
    • 3 years ago
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    First combine the logs ( addition of log is multiplication) log(x+4) + log(x-4) = Log ((x+4)/(x-4)) raise everything from base 9 for part one, doing this cancels the log because they are inverse functions and 9^1 is just 9 resulting in (x+4)(x-4) = 9 distribute x^2 -16 = 9 x^2 = 25 x= + or -5

  2. GregTheo
    • 3 years ago
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    typo, the log product should read log ((X+4)(x-4))

  3. sami-21
    • 3 years ago
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    ok lets solve it since \[\Large \log(a)+\log(b)=\log(ab)\] (addition changes to multiplication \[\Large \log _{9}((x+4)(x-4))=1\] since \[\Large a^2-b^2=(a+b)(a-b)\] so \[\Large \log _{9}(x^2-16)=1\]

  4. GOODMAN
    • 3 years ago
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    Yes, I got as far as @sami-21 But got stuck. I knew the answer was 5. Didnt know how to get it.

  5. sami-21
    • 3 years ago
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    so now we can cgange the log by change of bass formula

  6. sami-21
    • 3 years ago
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    change*

  7. GOODMAN
    • 3 years ago
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    Hmm..thats where i got stuck. How do you do that?

  8. sami-21
    • 3 years ago
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    \[\Large \log _{a}(b)=\frac{\ln(a)}{\ln{b}}\]

  9. sami-21
    • 3 years ago
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    it is upto use to change it to any base .i used natural log whose base is e

  10. GOODMAN
    • 3 years ago
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    Hmm...i think i get it.

  11. sami-21
    • 3 years ago
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    ok

  12. GOODMAN
    • 3 years ago
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    I had another one similar to it: 7^(5x-2)=5^(3-x)

  13. GOODMAN
    • 3 years ago
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    I know you have to log both sides.

  14. sami-21
    • 3 years ago
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    yes take ln of both sides

  15. sami-21
    • 3 years ago
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    then use property \[\Large \log _{a}(b)^n=nlog _{a}b\] which means exponents gets down and gets multiplied !

  16. sami-21
    • 3 years ago
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    let me do the your question now using above property

  17. sami-21
    • 3 years ago
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    take ln of both sides\[\Large \ln(7)^{5x-2}=\ln(5)^{3-x}\] so using the above mentioned property \[\Large (5x-2)\ln(7)=(3-x)\ln(5)\]

  18. sami-21
    • 3 years ago
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    now use calculator to find values of ln7 and ln5

  19. GOODMAN
    • 3 years ago
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    ln7=1.946 ln5=1.609

  20. sami-21
    • 3 years ago
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    ln(7)=1.94 ln(5)=1.60 (5x-2)*(1.69)=(3-x)1.609

  21. sami-21
    • 3 years ago
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    can you solve this !!

  22. GOODMAN
    • 3 years ago
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    Yesss. I can take it from there, thank you so much! And the last one, so sorry, is b on the question.

  23. sami-21
    • 3 years ago
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    a typo ! correct is (5x-2)*(1.94)=(3-x)1.609

  24. sami-21
    • 3 years ago
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    ok. u mean this \[\Large \sqrt{3x-2}-\sqrt{x}=1\] ??

  25. GOODMAN
    • 3 years ago
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    Correct :)

  26. sami-21
    • 3 years ago
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    ok taking square of both sides

  27. GOODMAN
    • 3 years ago
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    Waaiit.

  28. sami-21
    • 3 years ago
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    we have \[\Large (\sqrt{3x-2}-\sqrt{x})^2=1^2\] using \[\Large (a-b)^2=a^2+b^2-2ab\] we have \[\Large ((3x-2)+x-2\sqrt{3x-2}\sqrt{x})=1\]

  29. GOODMAN
    • 3 years ago
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    Sorry, it was 3x-3

  30. sami-21
    • 3 years ago
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    ok

  31. GOODMAN
    • 3 years ago
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    Itsokay i will pretend the 2 was really 3

  32. sami-21
    • 3 years ago
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    ok

  33. GOODMAN
    • 3 years ago
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    Sorry :(

  34. sami-21
    • 3 years ago
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    its ok :) i am notorious for typo mistakes ask anyone here :P

  35. sami-21
    • 3 years ago
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    have a look at here http://www.wolframalpha.com/input/?i=solve%28sqrt%283x-3%29-sqrt%28x%29%3D1%29 its quite difficult to type :P

  36. GOODMAN
    • 3 years ago
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    Wow, thats good enough! Thank you soo much for your help. I learned a lot!

  37. GOODMAN
    • 3 years ago
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    Also, thank you too @GregTheo

  38. sami-21
    • 3 years ago
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    yw:)

  39. muhammad9t5
    • 3 years ago
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    \[\sqrt(3x-2)-\sqrt(x)=1\] add \[\sqrt(x)\] in both sides

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