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Why (a log x = b log x / b log a)?

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  1. Feb 6, 2017 #1
    Could someone please explain to me why this equation holds:

    alog x = blog x / blog a

    For all the all other rules concerning logarithms I could derive them from the rules for exponentiation, except for this rule. Could someone please explain to me how this derivation works? And how the value of be can be practically determined so that it gives a neat answer for blog a.

    For example as applied to 3/2log 3 = blog 3 / blog 3/2

    How to determine the value of b in this case such that blog 3/2 gives a neat answer?

    Thanks in advance!
     
  2. jcsd
  3. Feb 6, 2017 #2

    FactChecker

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    Your notation is strange. The logarithm to the base b of x is denoted logb x.
    I assume you are asking why loga x = logb x/logb a

    logb x
    = logb x ( logb a / logb a)
    = logb a ( logb x / logb a)

    So
    x = a( logb x / logb a)
    loga x = ( logb x / logb a)
     
  4. Feb 6, 2017 #3

    Ray Vickson

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    ##y = \log_a(x)## means that ##y## is the solution of the equation ##a^y = x##. However, ##a = b^{\log_b a}##, so ##a^y = (b^{\log_b a})^y = b^{ y \log_b a} = x##, which means that ##\log_b x = y \log_b a = \log_a x \log_b a.##
     
  5. Feb 7, 2017 #4
    Okay, thanks for your answers! I get the derivation now. But what I am still unclear about is this:

    -Suppose you want to calculate y for loga (x) = y. When do you decide that the best way to do is by using the rule loga (x) = logb (x) / logb (a)? And how do you determine the value of b in such a case? Is b arbitrary? Could you please give a concrete example with actual numbers instead of just variables?

    I appreciate you answers very much. I'm trying to understand all this through self-study, without a teacher. So I guess that makes you my teachers :)

    As for my strange notation, I don't know if this is a European thing as opposed to an Anglo-American thing, but the book from which I am studying writes alog (x) instead of loga (x).

    Also, I am sorry for erasing the template. I had no idea it was obligatory. I just thought: 'What's all this stuff doing in my text box?' I thought I had hit a wrong button or something...
     
  6. Feb 7, 2017 #5

    ehild

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    Usually we use e or 10 for b. But b can be arbitrary.
    If you want to take the derivative of loga(x) transform it to base e logarithm, as the derivative of log(x) is simply 1/x.
    If you have an expression with logarithms with different bases, choose an appropriate one. and transform all the others.
     
  7. Feb 7, 2017 #6

    LCKurtz

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    To add to what others have said, sometimes you actually want to calculate a log to an unusual base. For example, searching a sorted table with N entries for a particular entry can be done by a computer by dividing the table in two, determining whether you have gone too far, then dividing the appropriate remaining half by two again and repeating until you find the entry you are looking for. The maximum number of steps needed to do that happens to be ##\log_2 N##. So how many searches are required if the table has 5000 entries? The answer is at most ##\log_2 5000##. But your handy calculator probably only does ##\log_{10}## or ##\log_e = \ln## keys. So you could calculate$$
    \log_2 5000 = \frac{\log_{10} 5000}{\log_{10} 2}$$with your calculator.
     
  8. Feb 13, 2017 #7
    it may be arbitrary with the caveat that b>0 and b is unequal to 1
    ##log_b(x)=y## and also x >0

    The common definition has logarithm and exponent functions being inverses of each other.

    Those belong to the definition of logarithm function which allow the logarithm and exponent functions to be inverse functions, I think.
    ##y=log_b(x)## <=> ##b^y=x##
     
  9. Feb 13, 2017 #8
    for example these were tricky but fun log problems that I just did a while ago...
    I'm also studying these logs for an exam in two weeks

    a)
    solve for k,
    ##log_k(25)=2##

    b) solve for k,
    ##log_k(2)=25##

    c) solve for x,
    ##log_2(x-3) +log_2(x-5)=3##

    d) for good measure an bonus one
    solve for x
    ##x*e^x+2e^x=0##

    e) very tricky one this. solve for x
    ##2e^{2x}+e^{x}-1=0##
     
  10. Feb 13, 2017 #9
    f) one more which is tricky... solve for x

    lg(x)+lg(x-21)=2
     
  11. Feb 14, 2017 #10

    I think vid was ok at Khan academy. For defiinition of log function you just have ##log_a(x)=y <=> a^y=x## and basenumber a>0,a≠1 and the numerus x>0
     
  12. Feb 14, 2017 #11

    LCKurtz

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    @late347: I'm glad you are enjoying studying logarithms. But have you noticed that you have been talking to yourself? The OP left the scene a week ago.
     
  13. Feb 17, 2017 #12
    Hi guys,

    No, I haven't left... Just been preoccupied with other stuff, but I'm studying all your answers right now... I think I got the logarithm thing under my belt now, thanks to all of you...

    Greetings,
    The OP :)
     
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