GR9768 Q 63
GR9768 Q 63
Ques : At how many points in the xyplane do the graphs of y = x^12 and y = 2^x intersect ?
A) None
B) 1
C) 2
D) 3
E) 4
I could figure out 2 points, but the ans is D.
Also any general method of solving such problems ?
Thanks.
A) None
B) 1
C) 2
D) 3
E) 4
I could figure out 2 points, but the ans is D.
Also any general method of solving such problems ?
Thanks.

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: GR9768 Q 63
since your ans is 2 , i guess you let x^12=2^x and use log_2 for both sides right? note that log is only defined for (0,+inf), looking back at the graph of the 2 func. you should see effortlessly that they intersect at some where when x<0.
Re: GR9768 Q 63
speedychaos4 wrote:since your ans is 2 , i guess you let x^12=2^x and use log_2 for both sides right? note that log is only defined for (0,+inf), looking back at the graph of the 2 func. you should see effortlessly that they intersect at some where when x<0.
No, I did not do it that way (by taking log)
I plotted the (rough) graphs.

 Posts: 27
 Joined: Tue Apr 06, 2010 8:22 am
Re: GR9768 Q 63
you may stop worrying right away, plotting graphs is the right thing to do. ETS just made a bloody typo, there is no third intersection point. for those who still unsure it is easy to go there > http://www.wolframalpha.com/input/?i=2^x+%3D+x^12 and see:
x ~= 0.9467803303936548
x ~= 1.063346830653734
just what one gets with rough plotting.
x ~= 0.9467803303936548
x ~= 1.063346830653734
just what one gets with rough plotting.

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: GR9768 Q 63
Sketching roughly might be what ETS wants to mislead you into.
Since every single question in the test is very important (at least true for me), you better be 90% or 100% sure before you choose the answer.
My approach is that, by letting
x^12=2^x,
we could have
log_2(x^12)=log_2(2^x),
which yields
log_2(x)=x/12
let
g(x)=log_2(x)x/12
then
g'(x)=1/(xln2)1/12
note that g'(x)>0 when x is close to 0 and there is a certain point x0 from which g'(x)<0
note also g(x)<0 when x is close to 0
from g'(x) we know that for x<x0, g(x) is strictly increasing and we can easily find a point at which g(x)>0, x=2 for example, and g(x) becomes strictly decreasing when x>x0, therefore, we could know that the 2 func. intersects twice on (0,+inf)
I afraid it is too complicated, but it works for me.
Since every single question in the test is very important (at least true for me), you better be 90% or 100% sure before you choose the answer.
My approach is that, by letting
x^12=2^x,
we could have
log_2(x^12)=log_2(2^x),
which yields
log_2(x)=x/12
let
g(x)=log_2(x)x/12
then
g'(x)=1/(xln2)1/12
note that g'(x)>0 when x is close to 0 and there is a certain point x0 from which g'(x)<0
note also g(x)<0 when x is close to 0
from g'(x) we know that for x<x0, g(x) is strictly increasing and we can easily find a point at which g(x)>0, x=2 for example, and g(x) becomes strictly decreasing when x>x0, therefore, we could know that the 2 func. intersects twice on (0,+inf)
I afraid it is too complicated, but it works for me.

 Posts: 27
 Joined: Tue Apr 06, 2010 8:22 am
Re: GR9768 Q 63
but you see that your conclusion is wrong since the equation has only one positive root (refer to wolfram alpha)speedychaos4 wrote:we could know that the 2 func. intersects twice on (0,+inf)

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: GR9768 Q 63
But if you take a more numerical approach, for example, by the wolfram figure you provide, it is clear that 2^x<x^12 when x=2, then take a large no, for example, x=120, then 2^x=2^120=(2^10)^12=1024^12>120^12=x^12, therefore they must intersects somewhere on (2,120)EugeneKudashev wrote:but you see that your conclusion is wrong since the equation has only one positive root (refer to wolfram alpha)speedychaos4 wrote:we could know that the 2 func. intersects twice on (0,+inf)

 Posts: 27
 Joined: Tue Apr 06, 2010 8:22 am
Re: GR9768 Q 63
the latter is pretty convincing, I lay down my arms
now the question is, why wolfram presents only two solutions (numerically). the third root is approx 74.6, as I was able to conclude via plotting the graphs in gnuplot. thanks anyway, up to this very day I was absolutely sure that it was a typo.
now the question is, why wolfram presents only two solutions (numerically). the third root is approx 74.6, as I was able to conclude via plotting the graphs in gnuplot. thanks anyway, up to this very day I was absolutely sure that it was a typo.
Re: GR9768 Q 63
Put more simply: an exponential eventually grows faster than any polynomial so 2^x will overtake x^12 at some large enough x.