Hard Logarithm Question

    • Thread Starter


(\log_3 p)^2 = \log_3(p^2)

\log_3(p+q)=\log_3 p + \log_3 q

    Solve these simultaneous equations.

    I've got  (\log_3 p)^2 = \log_3 (\frac{p^2 +qp}{q}) so far, but I don't know how to eliminate q.

    Solving the first equation gives you p, and then it's easy to find q from the 2nd.
    • Thread Starter

    Ok thanks, got it.

    (Original post by Dan12)
    How do you find p?
    In the first equation, for the RHS, use the logarithmic identity that states that "the logarithm of the n-th power of a number is the same as n times the logarithm of the number." Then, treat the entire equation as a quadratic in log_3p (hint: substitute m=log_3p). Let me know if you still don't get it.

    Q1. (log3P)2 = log3(P2)
    (log3P)2 = 2(log3P)

    Let log3P = x

    x2 = 2x
    x = 2

    log3P = 2
    P = 9

    Q2. log3(P+Q) = log3P + log3Q
    log3(P+Q) = log3(PQ)
    P+Q = PQ

    Substitute P = 9 from Q1

    9+Q = 9Q
    9 = 8Q
    9/8 = Q
    Q= 1 1/8 or 1.125

    Therefore, P is 9 and Q is 1.25

    (Original post by 2012_jonathan_s)
    Welcome to TSR Jonathan.

    Two rules you should know.

    Don't bump old threads for no reason.

    Don't provide full solutions.
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