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pratu043

  • 3 years ago

Can someone please explain how to solve a log equation with an example.

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  1. dpaInc
    • 3 years ago
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    how 'bout u pick an example....

  2. dpaInc
    • 3 years ago
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    otherwise i'll choose the easiest...:)

  3. pratu043
    • 3 years ago
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    OK ... \[5^{x - 3} . 3^{2x-8} = 225\]

  4. pratu043
    • 3 years ago
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    How to solve this using log? I'm new so I have no idea.

  5. pratu043
    • 3 years ago
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    I mean new to log.

  6. pratu043
    • 3 years ago
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    pls help.

  7. lgbasallote
    • 3 years ago
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    welcome to logarithms!

  8. pratu043
    • 3 years ago
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    Yeah thanks!

  9. waterineyes
    • 3 years ago
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    I guess you can solve this without using Logs..

  10. pratu043
    • 3 years ago
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    But I want to know how to do it with log.

  11. waterineyes
    • 3 years ago
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    Okay then all here will explain you how to do this...

  12. moha_10
    • 3 years ago
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    15^(x-3)+(2x-8)=225 now we take ln (3x-11)15= ln225

  13. pratu043
    • 3 years ago
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    What is In? And how did you get 15^(x-3)+(2x-8)=225?

  14. moha_10
    • 3 years ago
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    15*ln(3x-11)=ln 225

  15. moha_10
    • 3 years ago
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    by multi 3 into 5 to get 15 and the exponent it's added

  16. pratu043
    • 3 years ago
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    But how can you just multiply the base and add the exponents?

  17. waterineyes
    • 3 years ago
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    Use the following formulas of Logs: \[\large Log(a \times b) = Log(a) + Log(b)\] \[\large Log(a)^b = b.Log(a)\]

  18. moha_10
    • 3 years ago
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    before we take the logarithem

  19. moha_10
    • 3 years ago
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    it's from exponent experepties

  20. pratu043
    • 3 years ago
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    Why did you delete that @dpaInc?

  21. dpaInc
    • 3 years ago
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    because i don't wanna take away from these guys awesome explanations....:)

  22. waterineyes
    • 3 years ago
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    See, take Log both the sides you will get: \[\large Log(5^{x-3}.3^{2x-8}) = Log(225)\] Now use the formulas I have written above..

  23. waterineyes
    • 3 years ago
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    Use the first formula only...

  24. pratu043
    • 3 years ago
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    \[\log(5^{x-3}) . \log(3^{2x-8}) = \log 225 \]

  25. waterineyes
    • 3 years ago
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    Now Check my first formula carefully...

  26. moha_10
    • 3 years ago
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    @pratu043 trust take my answer

  27. moha_10
    • 3 years ago
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    like what waterineyes did now

  28. pratu043
    • 3 years ago
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    That's what I've done here right @waterineyes?

  29. waterineyes
    • 3 years ago
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    I give you an example: \[\large Log(2 \times 3) = Log(2) + Log(3)\] Go by this procedure...

  30. pratu043
    • 3 years ago
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    I've already done that. So now we should use the second one?

  31. waterineyes
    • 3 years ago
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    You have done it wrongly my friend...

  32. waterineyes
    • 3 years ago
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    See check the formula.. After formula the result will be in sum form and your result is not in sum form.. Is there any + sign in your result???

  33. pratu043
    • 3 years ago
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    Oops. \[\log(5^{x-3}) + \log(3^{2x-8}) = \log225\]

  34. waterineyes
    • 3 years ago
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    Yes now you are right... Now 225 is square of what number???

  35. pratu043
    • 3 years ago
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    15

  36. waterineyes
    • 3 years ago
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    So can you write 225 as \(15^2\) on right hand side.?? If yes then do it..

  37. pratu043
    • 3 years ago
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    \[\log(5^{x-3}) + \log(3^{2x-8}) = \log15^{2}\]

  38. waterineyes
    • 3 years ago
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    Now use the power rule of Logarithms that is the Second Formula that I have given above.. Can you use that??

  39. waterineyes
    • 3 years ago
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    I give you an example of that too: \[\large Log(x)^4 = 4.Log(x)\]

  40. pratu043
    • 3 years ago
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    \[(x-3)\log5 + (2x-8)\log3 = 2\log15\]

  41. waterineyes
    • 3 years ago
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    Can you write 15 as 3*5??

  42. pratu043
    • 3 years ago
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    Yes. \[(x-3)\log5 + (2x-8)\log3 = 2\log(5 * 3)\]

  43. waterineyes
    • 3 years ago
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    Then use it and use the first formula again on right hand side...

  44. pratu043
    • 3 years ago
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    \[(x-3)\log5 + (2x-8)\log3 = 2\log5 + 2\log3\]

  45. waterineyes
    • 3 years ago
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    Now what you have to do is to compare the coefficients of \(Log3\) and \(Log5\) both the sides can you do that??

  46. waterineyes
    • 3 years ago
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    Tell me the coefficient of Log3 on Left hand side and right hand side...

  47. pratu043
    • 3 years ago
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    On LHS its 2x - 8 and on RHS its 2.

  48. waterineyes
    • 3 years ago
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    Just equate them and find x from it..

  49. waterineyes
    • 3 years ago
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    Similarly equate the coefficients of log5 and then also find x you will get the same x in both the case...

  50. waterineyes
    • 3 years ago
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    No no...

  51. pratu043
    • 3 years ago
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    \[(2x-8)\log3 = 2\log3\]

  52. waterineyes
    • 3 years ago
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    I said to equate the coefficients buddy.. Coefficients are 2x - 8 and 2.. you should equate this only...

  53. pratu043
    • 3 years ago
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    So you should just do 2x-8 = 2?

  54. pratu043
    • 3 years ago
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    x = 5

  55. waterineyes
    • 3 years ago
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    See, if : \[\large xlog3 = ylog3\] implies x = y. Getting??

  56. pratu043
    • 3 years ago
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    Yes!!

  57. waterineyes
    • 3 years ago
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    Yes you have found x = 5 that is absolutely right...

  58. pratu043
    • 3 years ago
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    Thanks a lot!!

  59. waterineyes
    • 3 years ago
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    Now equate the coefficients of log5..

  60. pratu043
    • 3 years ago
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    x - 3 = 2 x = 2 + 3 = 5

  61. waterineyes
    • 3 years ago
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    Yes got the same value of x so you are right.. If anyhow x values are coming different then either question is wrong or your solution is wrong.. Getting??

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