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Thread: Logarithms

  1. #1
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    Logarithms

    Hi! I understand the concepts regarding logarithms, but I'm having some trouble showing that the following statement is true, if you could help me out I would really appreciate it 3/(log[2]a) - 2/(log[4]a) = 1/(log[0.5]a) *the square brackets indicate that the log is in the base of the number inside the brackets *THEY ARE NOT NATURAL LOGARITHMS* Thanks again!

  2. #2
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    Quote Originally Posted by teddy7 View Post
    Hi! I understand the concepts regarding logarithms, but I'm having some trouble showing that the following statement is true, if you could help me out I would really appreciate it 3/(log[2]a) - 2/(log[4]a) = 1/(log[0.5]a) *the square brackets indicate that the log is in the base of the number inside the brackets *THEY ARE NOT NATURAL LOGARITHMS* Thanks again!
    Hint:

    Assume Log2(a) = x → 2x = a

    calculate Log4(a) and Log0.5(a) as a function of 'x'
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    Hello, teddy7!

    Formula: .[tex]\log_b(a) \:=\:\dfrac{\ln a}{\ln b}[/tex]


    [tex]\displaystyle\text{Show that: }\:\frac{3}{\log_2(a)} - \frac{2}{\log_4(a)} \;=\;\frac{1}{\log_{0.5}(a)}[/tex]

    The left side is: .[tex]\dfrac{3}{\frac{\ln(a)}{\ln(2)}} - \dfrac{2}{\frac{\ln(a)}{\ln(4)}} \;\;=\;\; \dfrac{3\ln(2)}{\ln(a)} - \dfrac{2\ln(4)}{\ln(a)} \;\;=\;\;\dfrac{\ln(2^3)}{\ln(a)} - \dfrac{\ln(4^2)}{\ln(a)} [/tex]

    . . . . . . . . . [tex]=\;\;\dfrac{\ln(8)}{\ln(a)} - \dfrac{\ln(16)}{\ln(a)} \;\;=\;\; \dfrac{\ln(8) - \ln(16)}{\ln(a)} \;\;=\;\;\dfrac{\ln\left(\frac{8}{16}\right)}{\ln( a)} [/tex]

    . . . . . . . . . [tex]=\;\;\dfrac{\ln(0.5)}{\ln(a)} \;=\;\dfrac{1}{\frac{\ln(a)}{\ln(0.5)}} \;\;=\;\;\dfrac{1}{\log_{0.5}(a)} [/tex]

  4. #4
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    Quote Originally Posted by teddy7 View Post
    Hi! I understand the concepts regarding logarithms, but I'm having some trouble showing that the following statement is true, if you could help me out I would really appreciate it 3/(log[2]a) - 2/(log[4]a) = 1/(log[0.5]a) *the square brackets indicate that the log is in the base of the number inside the brackets *THEY ARE NOT NATURAL LOGARITHMS* Thanks again!
    Another way:

    Log2(a) = x → 2x = a → 4x/2 = a → Log4(a) = x/2

    Log2(a) = x → 2x = a → (1/2)-x = a → Log1/2(a) = -x

    Then

    3/Log2(a) - 2/Log4(a)

    = 3/x - 2/(x/2)

    = 3/x - 4/x

    = 1/(-x)

    = 1/Log1/2(a)
    Last edited by Subhotosh Khan; 08-10-2013 at 08:51 PM. Reason: fixed some inconsequential formatting issue
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #5
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    Regarding
    Quote Originally Posted by soroban View Post
    . . . . . . . . . [tex] \ \ ... \ =\;\;\dfrac{\ln(0.5)}{\ln(a)} \;=\;\dfrac{1}{\frac{\ln(a)}{\ln(0.5)}} \;\;=\;\;\dfrac{1}{\log_{0.5}(a)} \ \ [/tex]
    and
    Quote Originally Posted by Subhotosh Khan
    ... = 1/Log1/2(a)
    [tex] \ \ \ [/tex] the answer's form can also be streamlined to avoid the fractional form as:

    [tex]\dfrac{\ln(0.5)}{\ln(a)} \ = [/tex]

    [tex]\dfrac{\ln(2^{-1})}{\ln(a)} \ = [/tex]

    [tex] \dfrac{- \ln(2)}{\ln(a)} \ = [/tex]

    [tex]\boxed{- \log_a(2)}[/tex]





    Edit: Yes, it is the "required form," but it is a useful thing to know, especially if the OP were to be shown an alternate form in the future.
    Last edited by lookagain; 08-10-2013 at 08:26 PM.

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