Re: Real

From: Danilo <danilorj@xxxxxxxxx>
Date: Thu, 14 Dec 2006 18:08:57 EST

Ok!
f(x) = 5^x (1 + (2/5)^x - (4/5)^x - (3/5)^x) > 5^x (1 - (4/5)^x - (3/5)^x)
Now g_1(x) = 1 - (4/5)^x - (3/5)^x is increasing and g_1(2) = 0, so there are no solutions for x > 2.
My doubt is how can you affirm that there are no solutions for x>2 in this case.
For example for x=3 we have g_1(3)>0
and f(x)>5^x*g_1(x) and ??
.

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