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  • Sep 16, 2007, 04:16 PM
    SparkysGurl706
    Algebra 4
    How do I solve problems on antilogs??
  • Sep 21, 2007, 09:19 AM
    ebaines
    Do you understand logarithms? The antilog is nothing more than the inverse function of logs, which boils down to raising the base of the logarithm to the number.

    For example, using base 10, log(100) = 2. Hence the antilog of 2 is 100. Stated another way, th antilog of a = base^a, so the antilog of 2 is 10^2 =100.

    Since antilogs are just exponents, they have the following properties:

    antilog(a+b) = antilog(a) * antilog(b)
    antilog(a-b) = antilog(a)/antilog(b)
    antilog(n*a) = (antilog(a))^n
    Notice that these properties may look similar to logarithms, but they are "backwards."

    For example, in base 10, what is the antilog of 3.5?
    antilog(3.5) = antilog(0.5 + 3) = antilog(0.5) * antilog(3) = (10^0.5) * (10^3)= sqrt(10) * 1000 = 3.1623 * 1000 = 3162.3

    On your calculator you could have just found 10^3.5 to get the same result.

    Hope this helps.

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