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Homework Help: Evaluating/Solving Logarithmic expressions/equations

  1. Jun 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Evaluate the expression "log base 8 of 16"

    2. Relevant equations
    Don't know of any

    3. The attempt at a solution
    I dont know where to begin. I have read the textbook chapter on logarithms three times over, and I am now more thoroughly confused than ever. My textbooks seems to be skipping steps and citing the "Def. of Logarithm" as the reasoning.

    For the problem above, my textbook says to set the expression equal to x, and then magically gets 8^x=16. After this, it gets an answer of 4/3, without mentioning how. I assume it wants me to look at the problem and be able to see that 8^(4/3)=16, but I don't think I can do that with all problems. I am confused as to why you set the expression equal to x, and then how the expression is rearanged to get 8^x.

    Can someone please walk me through this problem without skipping any steps?
  2. jcsd
  3. Jun 27, 2010 #2


    Staff: Mentor

    This isn't magic. They are using the definition of the logarithm, which says:
    x = logay is equivalent to y = ax

    (There are some restrictions on y and a that I have omitted.)

    For your expression, you have x = log8 16, which is equivalent to 16 = 8x. There are no missing steps; the book is using the defining relationship between a log equation and the equivalent exponential equation.

    It's useful to think of the log, base a, of a number as the exponent on a that produces that number. So if x is the log, base 8, of 16, then 8x = 16.

    Continuing the problem requires knowledge of the laws of exponents, one of which is that (am)n = amn. 16 is a power of 2, and so is 8. Another property is that am = an ==> m = n.

    You have 16 = 8x, so 24 = (23)x.

    Can you finish the problem now?

  4. Jun 27, 2010 #3
    Thanks, I see how to get 4/3 now.

    What about this problem?

    log base 6 of x = (.5 log base 6 of 9)+(.33 log base 6 of 27)

    How do I get rid of the logs? Since they have the same base, can I just divide all the terms by log base 6 to cancel everything out?
  5. Jun 27, 2010 #4


    Staff: Mentor

    Absolutely not! log6 is not a number -- it's a function

    What properties of logs (any base) do you know? LIst the ones you know and we can go from there.
  6. Jun 27, 2010 #5
    Well, I think this is the one applicable to this problem:

    Product: log base b of mn = log base b of m + log base b of n

    What do I do with the numbers in front of the logs though?
  7. Jun 27, 2010 #6


    Staff: Mentor

    Another one is logb an = n logba
  8. Jun 27, 2010 #7
    Ok, thats what I needed to solve this problem. Thanks!
  9. Jun 28, 2010 #8
    Remember not to exponentiate unless you have only one log on one or both sides of the equal sign. Dunno why, but I always used to forget about that...
  10. Jun 29, 2010 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You can exponentiate whenever you want as long as you keep all the properties of exponentiation and logarithms straight
  11. Jul 1, 2010 #10
    The difficulty in this prompts many to advise against exponentiating until exactly one log expression is on each side of the equality. But yeah, you can...
  12. Jul 4, 2010 #11
    Another solution is to raise both sides to a power of 6.

    6 ^ (log base 6 of x) = 6 ^ [(.5 log base 6 of 9)+(.33 log base 6 of 27)]
    Since 6^x and log base 6 of x are inverse functions, then
    x = [ 6 ^ (.5 log base 6 of 9) ] [ 6 ^ (.33 log base 6 of 27) ]

    Using the Power Rule for Logs ..

    x = [ 9 ^ 0.5 ] [ 27 ^ 0.33 ]
    x = 3 x 3
    x = 9
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