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Exponential Functions

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Show that if
log_b (a) = c, and log_y (b) = c, then log_a (y)=c^-2

  • Exponential Functions -

    logb a = log a/log b (base 10, or any other base for that matter)

    so logb(a) = loga/logb = c
    and logy(b) = logb/logy = c

    then [loga/logb][logb/logy] = (c)(c) = c^2
    so loga/logy = c^2 or
    logy/loga = 1/c^2 = c^-2

    but logy/loga = loga(y)

    so loga(y) = c^-2 as required.

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