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November 23, 2014

November 23, 2014

Posted by **Anonymous** on Tuesday, June 12, 2012 at 2:58am.

- math -
**Reiny**, Tuesday, June 12, 2012 at 8:46amThe range of any standard log function like yours is the set of real numbers.

That is, choosing the appropriate value of x will yield any y value you want

suppose we want y or f(x) to be -500

-500 = log(x-3)

x-3 = 10^-500

x = 3 + 10^-500 , a number just a bit to the right of +3

(the closer you get to +3 from the right, the larger into the negatives the y will become)

try the same thing for large values of x

e.g. x = 1 000 000 000

y = log (1 000 000 000 - 3) = 8.999..

to get really large values of y as an answer your x must be un-measurably large.

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