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anonymous

  • 5 years ago

Solve the logarithmic equation for x: log base 2(x+1)+log base 2(x-1)=3

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  1. nowhereman
    • 5 years ago
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    Apply exponentiation!

  2. anonymous
    • 5 years ago
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    .........huh? :P

  3. nowhereman
    • 5 years ago
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    If you know the logarithm you should know the exponential.

  4. anonymous
    • 5 years ago
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    I'm lost........ Gah why does math have to be so hard?

  5. nowhereman
    • 5 years ago
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    It's not, if you follow the definitions. How did you define log?

  6. anonymous
    • 5 years ago
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    What do you mean? I didn't define anything... I don't think...

  7. nowhereman
    • 5 years ago
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    I mean in class, or where you learned about the logarithm. Or how do you want to solve any problem containing it, if you don't know what it is?

  8. anonymous
    • 5 years ago
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    I learned this like 6 months ago, so I really don't remember. This is a take home test... I just need to solve for x, I think.

  9. nowhereman
    • 5 years ago
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    Then you should look it up, in your notes actually, but wikipedia is often good too. What you need here is \[b^{\log_b x} = x\] because logarithm is the inverse function of the exponential.

  10. anonymous
    • 5 years ago
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    Is b the base?

  11. nowhereman
    • 5 years ago
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    Indeed

  12. anonymous
    • 5 years ago
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    So it would be \[2^\log2(3)=3\]?

  13. watchmath
    • 5 years ago
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    hi nowhere man, we need to combine the logs first before applying exponentiation. Also in US usually they don't teach exponentiation explicitly in that way.

  14. anonymous
    • 5 years ago
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    That's what I thought, cause I didn't remember anything like that.... Help?

  15. watchmath
    • 5 years ago
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    Do you remember this property? \(log_b A+\log_b B= \log_b(AB)?\)

  16. anonymous
    • 5 years ago
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    I think so...

  17. watchmath
    • 5 years ago
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    Ok, what do you get if you apply that property to the left hand side of your equation?

  18. anonymous
    • 5 years ago
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    \[\log_{2} 2x\]I think. Because the +1 and the -1 would cancel out.

  19. watchmath
    • 5 years ago
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    But you multiply the inside instead of adding them.

  20. anonymous
    • 5 years ago
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    Why would I multiply them? It says I'm adding the two logs...

  21. anonymous
    • 5 years ago
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    Answer x = 3; log(3+1) to the base 2 = 2 and log(3-1) to the base 2 = 1

  22. watchmath
    • 5 years ago
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    look at again \(log_b A +\log_b B=\log_b(AB)\) In there you multiply A and B

  23. anonymous
    • 5 years ago
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    Ohhhhh okay. So it would be\[\log_{2} x^2+1\]

  24. watchmath
    • 5 years ago
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    Check it again what do you get if you do (x+1)(x-1) (foil it :) )

  25. anonymous
    • 5 years ago
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    Formula log(A) to the base 2 + log(B) to the base 2 = log(A B) to the base 2; Steps: 1. log (x+1)(x-1) to the base 2 = 3 2. removing log, 2 raised to 3 = (x+1) (x-1) 3. 8 = x^2 - 1 4. 9 = x^2 5. taking square root on both sides, 3 = x

  26. anonymous
    • 5 years ago
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    \[\log_{2} x^2-1\]

  27. anonymous
    • 5 years ago
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    that would equal 3, right?

  28. watchmath
    • 5 years ago
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    great! So now we have \(\log_2(x^2-1)=3\) Do you remember how to change from log equation into exponential equation?

  29. anonymous
    • 5 years ago
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    I think if I make the three on the other side a log as well I could do it, but that doesn't seem right...

  30. watchmath
    • 5 years ago
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    Remember this \[\log_bx=y\iff b^y=x\]

  31. anonymous
    • 5 years ago
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    That's right

  32. anonymous
    • 5 years ago
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    OHH I REMEMBER THAT! :D

  33. anonymous
    • 5 years ago
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    So it would be \[2^3=x^2-1\]

  34. anonymous
    • 5 years ago
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    Yep. Now solve for x :)

  35. anonymous
    • 5 years ago
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    \[x^2=9\]so x=3 :D

  36. anonymous
    • 5 years ago
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    AHHH YOU GUYS ARE AMAZING<3

  37. anonymous
    • 5 years ago
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    x could be -3 as well. Since we are dealing with logarithms, we ignore -3

  38. anonymous
    • 5 years ago
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    Oh yeah, I forgot about that. Negatives don't work with logs.

  39. watchmath
    • 5 years ago
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    To bee a little pedantic from \(x^2=9\) we have \(x=\pm 3\). But when we plug in back x=-3 into the equation we are in trouble because we can't compute log of negative number. SO x=3 is the only solution.

  40. anonymous
    • 5 years ago
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    YAYAYAYYAYAYAYAYYAYAY!<3 Thanks so much guys!

  41. anonymous
    • 5 years ago
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    Any time!

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