# quick C2 questionWatch

This discussion is closed.
#1
Brain has been frazzled by stats...
prove that the intersection point of y=2*2^x and y=2^-2 +1 is (0,2)
thanks
0
13 years ago
#2
y = 2*(2^x) = 2^(x+1)
y = 2^-2 + 1 = 1.25

1.25 = 2^(x+1)
log 1.25 = log (2^(x+1))
log 1.25 = (x+1)log 2
x + 1 = log 1.25 / log 2
x = log 1.25 / log 2 - 1
x = -0.678071905112637652129680570510 61

Therefore PoI(-0.678,1.25)

Which ain't right?
0
13 years ago
#3
(Original post by franks)
Brain has been frazzled by stats...
prove that the intersection point of y=2*2^x and y=2^-2 +1 is (0,2)
thanks
subsitutite y = 2 x 2^x into equation:

2 x 2^x = 2^-2x + 1
2^(x+1) = 2^-2x + 1
0 = 2^-2x - 2^(x+1) + 1

sub (x=0) > 2^-2(0) - 2^(0+1) +1 = 2

Hence line intersects other line when x = 0, y = 2
0
13 years ago
#4
(Original post by TomX)
y = 2*(2^x) = 2^(x+1)
y = 2^-2 + 1 = 1.25

1.25 = 2^(x+1)
log 1.25 = log (2^(x+1))
log 1.25 = (x+1)log 2
x + 1 = log 1.25 / log 2
x = log 1.25 / log 2 - 1
x = -0.678071905112637652129680570510 61

Therefore PoI(-0.678,1.25)

Which ain't right?
Yeh I got that before, but he made a mistake and missed an x in the power of 2
0
13 years ago
#5
(Original post by Vijay1)
Yeh I got that before, but he made a mistake and missed an x in the power of 2
I thought he probably had missed an X somewhere, since it was strange y just equalled a constant, which was wrong.
0
#6
subsitutite y = 2 x 2^x into equation:

2 x 2^x = 2^-2x + 1
2^(x+1) = 2^-2x + 1
0 = 2^-2x - 2^(x+1) + 1

sub (x=0) > 2^-2(0) - 2^(0+1) +1 = 2

Hence line intersects other line when x = 0, y = 2
ahh so there's no way of doing it without subbing 0 in? I was trying all sorts of wierd things !
Thanks !
0
13 years ago
#7
is that proof? bad wording of the question, surely they should use "show". how many marks is it?
0
13 years ago
#8
Yeah that's the proof they're looking for, probably would say "show" in an exam.

That's c1 non?
0
13 years ago
#9
(Original post by Chris.)
Yeah that's the proof they're looking for, probably would say "show" in an exam.

That's c1 non?
hmm, its could be c2 or c1.
0
#10
well the rest of the question was to do with logs, so it was C2. Is there any way of working it out using logs ??
Thanks for the help!
franks x
0
13 years ago
#11
(Original post by franks)
well the rest of the question was to do with logs, so it was C2. Is there any way of working it out using logs ??
Thanks for the help!
franks x
There will be, can you write down the actual question, make you get it all right this time too :P
0
13 years ago
#12
(Original post by Vijay1)
sub (x=0) > 2^-2(0) - 2^(0+1) +1 = 2
Err, let's simplify that:

2^-2(0) - 2^(0+1) +1 =
2^0 - 2^1 + 1 =
1 - 2 + 1 = 0

This implies the PoI is (0,0), so I don't think he missed a two out of the power. Let's wait until he tells us the proper question.
0
13 years ago
#13
2 * 2^x = 2^-2x + 1 is the question, i can get as far as:

2^(x+1) = 1/2^(2x) + 1
2^(2x)*2^(x+1) = 1 + 2^(2x)
2^(3x+1) = 1 + 2^(2x)
2^(3x+1) = 1 + 2^(2x)
(3x+1)*log(2) = log(1 + 2^(2x))
:|
0
#14
lol im a 'she' by the way - i didnt realise you were referring to me about missing out a power of 2.... I will check again, here is the whole question:
lol no i put in an x instead of a 2 - sooooooooryyy ! - so what you decided the question was wasnt exactly right....

prove that the point of intersection of the currves y=2*2^x and y=2^-x +1 is (0,2)

thank you, you are all stars - sorry im such a twit,

franks, x
(female!)
0
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