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Logarithmic Problem

  1. Sep 8, 2014 #1
    Suppose your pocket calculator is damaged: it can only evaluate logarithms of numbers between 0 and 1. Find a way to evaluate the following logarithms with the use of such a calculator.

    log2 = log (10*0.2) = log10 + log0.2 = 1 + log0.2

    log333 = log(1000 * 0.333) = log10^3 + log0.333 = 3 + log0.333

    log1.1 = log(10 * 0.11) = log10 + log0.11 = 1 + log0.11

    log7588.56 = log(10 000 * 0.758856) = log10^4 + log0.758856 = 4 + log0.758856

    Are these solutions correct?
     
  2. jcsd
  3. Sep 8, 2014 #2

    jedishrfu

    Staff: Mentor

    They look okay.

    Don't you still have to look up the log(0.333) to complete the answer for 3 + log0.333 as an example?
     
  4. Sep 8, 2014 #3

    Mark44

    Staff: Mentor

    I believe the goal of the exercise is to reduce log expressions to a form for which they can be calculated by the defective calculator, but not to actually do the calculation.
     
  5. Sep 8, 2014 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Once upon a time- in the years "B.C." (Before Calculators) it was common to look up logarithms in tables- which only gave the logarithms for 0 to 1. To find the logarithm of a number such as 7588.56, yes, you would write it as [itex]0.758856 \times 10^{3}[/itex] and then [itex]log(0.758856)= log(0.758856 \times 10^4)= 4+ log(0.758856)[/itex].
     
    Last edited: Sep 8, 2014
  6. Sep 8, 2014 #5

    Mark44

    Staff: Mentor

    That last line should be
    ##log(7588.56)= log(0.758856 \times 10^4)= 4+ log(0.758856)##
     
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