TestMagic Forums - View Single Post

Dragonfinity · 2005 February 17th, 03:33 PM

NO. The graphs of y = x ^12 and y = 2 ^x intersect twice over the positive x-axis and once over the negative x-axis. Draw them again more carefully. I am not going to take the time to give an exact solution because it is a real pain but note the following:

x=1: x^12 = 1, 2^x = 2, hence x^12 < 2^x
x=2: x^12 = 4096, 2^x = 4, hence x^12 > 2^x, so there is clearly one pt. of intersection
The following is an approximation by computer, but the missing digits are irrelevant, as you will see:
x=100: x^12 = 1E+24, 2^x = 1.27E+30, hence x^12 < 2^x, so there must be a second pt. of intersection.

Does this end it? Or do I need to actually work out the problem! :-)

2005 February 17th, 03:33 PM	#11 (permalink)
Dragonfinity Socratic realist Join Date: Feb 2005 Location: Georgia, USA Posts: 220	Re: Intersection NO. The graphs of y = x ^12 and y = 2 ^x intersect twice over the positive x-axis and once over the negative x-axis. Draw them again more carefully. I am not going to take the time to give an exact solution because it is a real pain but note the following: x=1: x^12 = 1, 2^x = 2, hence x^12 < 2^x x=2: x^12 = 4096, 2^x = 4, hence x^12 > 2^x, so there is clearly one pt. of intersection The following is an approximation by computer, but the missing digits are irrelevant, as you will see: x=100: x^12 = 1E+24, 2^x = 1.27E+30, hence x^12 < 2^x, so there must be a second pt. of intersection. Does this end it? Or do I need to actually work out the problem! :-)