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Math Help - Help with logs please

  1. #1
    Newbie
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    May 2005
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    Help with logs please

    We just started doing some work on logs and i cant work out this problem, i think its quite easy but im crap at these.

    Two curves:

    y=a^x and y=2b^x

    Prove that x coord of the point of intersection is 1 / (log(of base2)A-log(of base2)B)

    Help would be appreciated
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  2. #2
    MHF Contributor
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    y=a^x and y=2b^x

    y = a^x ----(1)
    y = 2(b^x) ----(2)

    At their point of intersection, the two logarithmic curves have the same coordinates.
    So, y from (1) = y from (2)
    a^x = 2(b^x) ----(3)

    Since we are looking for the x-coordinate of the intersection, then we solve for x.
    Take the logs, (log to the base 10), of both sides of (3),
    x*log(a) = log(2) +x*log(b)
    Collect the x-terms,
    x*log(a) -x*log(b) = log(2)
    x[log(a) -log(b)] = log(2)
    Divide both sides by [log(a) -log(b)],
    x = log(2) / [log(a) -log(b)] -----(4)

    (4) is on base 10.
    Since you want x in base 2, then we transform the logs in 4 into logs to the base 2.

    You still remember how to change bases for logarithms?
    log(base a) to log(base b).

    log(base a)(N) = [log(base b)(N)] / [log(base b)(a)] ---***

    So,
    >>>log(base 10)(2)
    = [log(base 2)(2)] / [log(base 2)(10)]
    = 1 / [log(base 2)(10)]

    >>>log(base 10)(a)
    = [log(base 2)(a)] / [log(base 2)(10)]

    >>>log(base 10)(b)
    = [log(base 2)(b)] / [log(base 2)(10)]


    Substitute those into (4),

    x = log(2) / [log(a) -log(b)] -----(4)

    x = log(base 10)(2) / [log(base 10)(a) -log(base 10)(b)]

    x = {1 / log(base 2)(10)} / {[log(base 2)(a) / log(base 2)(10)] -[log(base 2)(b) / log(base 2)(10)]}

    Combine the denominator into one fraction only,
    x = {1 / log(base 2)(10)} / {[log(base 2)(a) -log(base 2)(b)] / log(base 2)(10)}

    Reverse the denominator, to multiply it to the numerator,
    x = {1 / log(base 2)(10)} * {log(base 2)(10) / [log(base 2)(a) -log(base 2)(b)]}

    The log(base 2)(10) cancels out,
    x = 1 / [log(base 2)(a) -log(base 2)(b)] ----answer.

    ===========
    Or, from (3), we can take the logs to the base 2 directly----without passing through the logs to the base 10.

    a^x = 2(b^x) ----(3)
    log(base 2)(a^x) = log(base 2)(2 b^x)
    x*log(base 2)(a) = log(base 2)(2) +x*log(base 2)(b)
    x*log(base 2)(a) = 1 +x*log(base 2)(b)
    x*log(base 2)(a) -x*log(base 2)(b) = 1
    x[log(base 2)(a) -log(base 2)(b)] = 1
    x = 1 / [log(base 2)(a) -log(base 2)(b)] ----answer.
    Last edited by ticbol; May 14th 2005 at 08:27 PM.
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  3. #3
    Newbie
    Joined
    May 2005
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    Ah ok!

    Thanks very much for your help
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