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Math Help - log change formula

  1. #1
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    Smile log change formula

    Can someone please help me with these optimisation questions finding it very difficult (ps have given what i think is valid answer to q1 but cant do q2 at all.)
    note: 2^6 measn 2 to the power of 6,
    note: theta is a capital theta
    note: log c (a) means log to the base c of a

    1) Let a, b, c be positive real numbers. Show that loga(b) = logc(b)/ logc(a).

    ( i think i have the answer to this can someone pls confirm this is a thoroguh enough proof

    let y= log a (b)
    then a^y =b
    so taking logarithms to the base c gives

    log c (a^y)= log c (b)
    implying
    ylog c(a) = log c (b)
    so: y = log c (b)/log c (a)
    giving answer

    2) Let a, b be real numbers, with a>0 and b > 0. Show (n + a)^b = theta(n^b). Show that (log n)^b = o(n^a). (For concreteness let’s take ‘log’ to mean ‘log2’.

    thanx edgar
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  2. #2
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    Quote Originally Posted by edgar davids View Post
    Can someone please help me with these optimisation questions finding it very difficult (ps have given what i think is valid answer to q1 but cant do q2 at all.)
    note: 2^6 measn 2 to the power of 6,
    note: theta is a capital theta
    note: log c (a) means log to the base c of a

    1) Let a, b, c be positive real numbers. Show that loga(b) = logc(b)/ logc(a).
    (It does not work if some are equal to 1).

    \log_a b = x, \ \log_c b = y , \ \log_c a = z .
    a^x = b , \ c^y = b , \ c^z = a
    a^x = b \implies (c^z)^x = b \implies c^{xz} = c^y \implies xz = y.
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