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Re: Logs
Posted:
Oct 14, 2013 11:55 PM


On Mon, 14 Oct 2013, Virgil wrote: > William Elliot <marsh@panix.com> wrote: > > On Mon, 14 Oct 2013, gyansorova@gmail.com wrote: > > > > > If x and y are both positive can we ever approximate > > > log(x+y) := log(x) + log(y) > > > > > x + y = e^log(x + y) = e^(log x + log y) = e^log x * e^log y = xy. > > Can you ever approximate xy = x + y? > > For x =/= 1, one can find y exactly, and for y =/= 1, > one can find x exactly, no approximating needed
0 = x  xy + y = (x  1)(y  1) + 1



