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Logarithm Question

  • 13-03-2008 09:06PM
    #1
    Smegball
    Registered Users, Registered Users 2 Posts: 2,283 ✭✭✭


    I'm sure this is very easy, but help would me greatly appreciated.
    2 log3x - log3 (x-2) = 2

    Could you solve this please and show workings.. would greatly help me thanxs :)


Comments

  • #2
    MathsManiac
    Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    I'm assuming both of those 3's are the base of the logs. If that's so, then you can use these log rules / properties:

    2log(a) = log (a^2), and

    log(a) - log (b) = log (a/b).

    Using both of these on the left hand side you get

    log[base3] (x^2 / (x-2)) = 2.

    Next use the fact that if log[base b] (a) = c, then b^c=a; (def. of log)

    So, you get 3^2 = x^2 / (x-2)

    So 9(x-2) = x^2,

    So x^2 -9x +18 = 0,

    So (x - 6)(x - 3) = 0, so x=6 or x=3.

    You should always check the answers to log-based equations, as sometimes you get "extraneous" roots, which don't actually work in the original equation, (they might give rise to logs of negative numbers).

    In this case, both roots work.


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