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Question about Log Graphs

  1. Oct 4, 2008 #1
    Can someone explain to me why the graph of:

    logy=logx² is the graph of a regular parabola

    logy=2logx is the graph of half a parabola (x>0)

    logxy=2 is the graph of half a parabola except x>0 and x cannot be equal to 1

    I just don't understand why they're not all normal parabolas, and how the second two are different than the first.
  2. jcsd
  3. Oct 5, 2008 #2
    (1) [itex]\log y = \log x^2[/itex] is a parabola.

    (2) [itex]\log y = 2\log x[/itex] is equivalent to (1) only when x > 0, because otherwise the logarithm does not exist. That is, whenever x > 0, (1) = (2), but when x < 0, we actually have [itex]\log y = 2\log(-x)[/itex].

    (3) [itex]\log_x y = 2[/itex]. You get this from (2) by dividing by [itex]\log x[/itex]. Thus, we already see that x > 0 from (2). Moreover, if x = 1, then [itex]\log 1 = 0[/itex], and you are dividing by 0, which is not allowed. Indeed, from (3), you see that if x = 1, then (3) can never equal 2 (1 to any power is still 1).
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