How do i solve this logarithmic expression?!-10 points (maths)?

If log4(x)=p, show that log2(x) =2p, well this is fairly simple and ive accomplished it. The problem is the next part.

Hence, find a) the value of k if log4(k) = 2+ log2(k) and b) the value of n if log2(n) + log4(n) =9

Well this is what i did==> log4(k) -log2(k) = 16====> log4(k)/ log2(k) =16 ....what now! Please help withh method and explanation!

Answer both questions for 10 points!

2 Answers

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  • iceman
    Lv 7
    9 years ago
    Favorite Answer

    Good job on your representation of log(base), i.e. log4(x), you may also see the following notations:

    log_4(x) = p , log_2(x) = 2p, or log₄(x) = p , log₂(x) = 2p

    solution: change the bases to either 10 or e(natural log):

    log₂(x) = log(x)/log(2)

    log₄(x) = log(x)/log(4) = p

    log(x)/log(2²) = p

    log(x)/2log(2) = p

    log(x)/log(2) = 2p = log₂(x)

    a)

    log₄(k) = 2 + log₂(k)

    log(k)/log(4) - log(k)/log(2) = 2

    [log(k) - 2log(k)]/ [2log(2)] = 2

    -log(k) = 2 * 2log(2)

    log(2^4) + log(k) = 0

    log(16k) = 0

    16k = 10^0 = 1

    k = 1/16

    All the best.

  • 9 years ago

    Let log k to the base 2 = a.

    So k = 2^a

    Let log k to the base 4 = b

    So k = 4^b = (2^2)^b = 2^(2b)

    Therefore 2^a = 2^(2b)

    and so a = 2b

    Given that b = 2 + a

    Hence a = 2(2 + a), a = -4, b = -2, k = 4^(-2) = 1 / 4^2 = 1/16

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