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Homework Help: Logarithmic question-

  1. Aug 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi, it maybe stupid question, but I struggle on this problem..

    2. Relevant equations

    Using properties:
    logabn=n logab
    logn(bc) = lognb+lognc

    3. The attempt at a solution
    I can't find the b number.
    log16(4354)+3/log25(43)=6 log162+4log165+3/3log254=6(log22/log216)+4(log25/log216)+1/log254=3/2 + 2log252= 3/2 + 4log25

    so a=3/2 b=4 but in the answer they find b to be 2 not 4
    Can you please tell me where is my mistake?

    Last edited: Aug 3, 2013
  2. jcsd
  3. Aug 3, 2013 #2


    Staff: Mentor

    You're missing a right parenthesis, so it's hard to tell what you're starting with.

    Did you mean log16(4354) +3/log25(43)?
  4. Aug 3, 2013 #3
    Lol, yes, Sorry it is definitely log16(4354)+3/log25(43)
  5. Aug 3, 2013 #4


    User Avatar
    Science Advisor

    Can we at least presume you know the basic properties of logarithms and so know that [itex]log_{16}(4^35^4)= 3log_{16}(4)+ 4 log_{16}(5)[/itex]? And, of course, since [itex]16= 4^2[/itex], [itex]4= 16^{1/2}[/itex] so that [itex]log_{16}(4)= 1/2[/itex]. That is, [tex]log_{16}(4^35^4)= 3+ 4log_{16}(5)[/tex]
    [itex]log_{25}(4^3)= 3log_{25}(4)[/itex].

    Perhaps what you are missing is that [itex]\log_a(x)= \frac{log_b(x)}{log_a(b)}[/itex]. You can use that to change the [itex]log_{25}[/itex] to [itex]log_{16}[/itex]. Further, since, [itex]16= 2^4[/itex], [itex]log_{2}(16)= 4[/itex] and so [itex]log_{16}(x)= \frac{log_2(x)}{4}[/itex]

    You can use that to change everything to [itex]log_2[/itex] as you have on the right.
    Last edited by a moderator: Aug 3, 2013
  6. Aug 13, 2013 #5
    I'm noob, but I think that you meant
    [itex]\log_a(x)= \frac{log_b(x)}{log_b(a)}[/itex]
    Mainly because if you start with one base and end with two it was not a change of base.
    I may be wrong.
  7. Aug 13, 2013 #6
    Don't worry, you're correct :tongue:
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