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Homework Help: Log questions

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve for X

    logx-log(x+11) = -1


    log4x-log4(x+15) = -1

    2. Relevant equations

    3. The attempt at a solution
    log x - log (x+11) = -1
    log (x/x+11) = -1

    I don't know how to solve for X after this point

    log4x-log4(x+15) = -1
    log4 x/(x+15) = -1

    I don't know how to get the X out of the log to solve for X
    Last edited: Oct 11, 2014
  2. jcsd
  3. Oct 11, 2014 #2


    Staff: Mentor

    The relationship you need for both problems is this one:
    loga(x) = y is equivalent to x = ay.

    For your first problem, log means log10 (lob base 10).
  4. Oct 11, 2014 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    What "base" of logs is used in the first question?

    Anyway, you certainly cannot have what you wrote, which was
    [tex] \log \left( \frac{x}{x} + 11 \right) = -1[/tex]
    which gives ##\log(12) = -1##. Did you mean
    [tex] \log\left( \frac{x}{x+11} \right) = -1?[/tex]
    If so, use parentheses, like this: log(x/(x+11)) = -1. At his point it matters what base you are using for log.

    For a given base ##b##, what number, ##y##, has ##\log_b(y) = -1##? Think about what that actually means.

    BTW: either use X or x, but not both in the same problem.
  5. Oct 11, 2014 #4
    I figured them out, thank you.

    log x - log (x+11) = -1
    log (x / (x+11)) = -1
    x/(x+11) = 10-1
    x/(x+11) = 0.1

    log4x-log4(x+15) =-1
    log4(x/(x+15) = -1
    x/(x+15)= 4-1
    x/(x+15) = 0.25
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