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Logarithim Problem I been stuck on for a day

  1. Jun 15, 2015 #1
    1. The problem statement, all variables and given/known data
    I have been trying to solve this problem all day, I have used up 3 pages and to no avail.
    I don't think i should try and change the bases because of the way the book is set up changing bases is not until the next chapter. So I want to try and get it out without changing bases. Can someone give me a hint?


    2433yp2.jpg

    The question is number 8
    2. Relevant equations


    3. The attempt at a solution
    My work is 3 pages of scratch work so it is hard to type out but all I have done is try to combine the logs shown into 2 separate logs for the base a and the base 10. I get stuck there since I can't remove the logs.
     
  2. jcsd
  3. Jun 15, 2015 #2

    pasmith

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    If the left hand side is [itex]\log_{a}(A) - \log_{10}(B)[/itex], what values do you have for [itex]A[/itex] and [itex]B[/itex]?
     
  4. Jun 15, 2015 #3
    I got it down to A being (6/25) and B being 24
     
  5. Jun 15, 2015 #4

    pasmith

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    You have simplified too far; consider [tex]\log_a (6/25) - \log_{10}(24) = \log_a(24/100) - \log_{10}(24) = \log_a(24) - 2\log_a(10) - \log_{10}(24).[/tex]
     
  6. Jun 15, 2015 #5
    I'm still lost because the bases are different.
     
  7. Jun 15, 2015 #6
    You have $$\log_{a}(6/25)=\log_{10}24-2$$
    Lets consider the -2 in the equation. What is ##\log_{10}0.01## equal to?

    Chet
     
  8. Jun 15, 2015 #7
    -2 is equal to log(0.01) [ I'm using Log for the base a , because I don't know Latex]

    so Log(24) - 2Log(10) = log(0.01) + log 24
     
  9. Jun 15, 2015 #8
    Now apply the product rule to log(0.01) + log (24). What do you get?
     
  10. Jun 15, 2015 #9
    so Log(24) - 2Log(10) = log(0.24)
     
  11. Jun 15, 2015 #10
    Now express 0.24 as a fraction reduced to lowest terms.

    Chet
     
  12. Jun 15, 2015 #11
    Log(24)- 2Log(10) = log(6/25)
     
  13. Jun 15, 2015 #12
    So, now you have $$\log_{a}(6/25)=\log_{10}(6/25)$$
    So, what is a equal to?

    Chet
     
  14. Jun 15, 2015 #13
    Oh I see it was just something simple as that . I was thinking I would have to remove the log in the end never thought that I would just have to compare the bases....
    So it was really easy I was overthinking it.
    Thank you so much .

    a is 10.
     
  15. Jun 15, 2015 #14

    Ray Vickson

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    Convert everything to ##\log_{10}##. There are standard formulas for converting logarithms between different bases.
     
  16. Jun 15, 2015 #15

    SammyS

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    OP said he was avoiding using that.
     
  17. Jun 16, 2015 #16

    Ray Vickson

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    Right: I missed that.
     
  18. Jun 16, 2015 #17

    SammyS

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    Problem (8) is equivalent to solving ##\ \log_a(A)=\log_{10}(A) \ ## for a.

    Suppose it had been equivalent to solve ##\ \log_a(A)=\log_{10}(A) \ ## for a, where A ≠ B . How would you do that without the change of base formula?
     
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