Mathematics GRE

Ques : At how many points in the xy-plane 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.

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.

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.

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.

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.

speedychaos4 wrote:we could know that the 2 func. intersects twice on (0,+inf)

but you see that your conclusion is wrong since the equation has only one positive root (refer to wolfram alpha)

EugeneKudashev wrote:

speedychaos4 wrote:we could know that the 2 func. intersects twice on (0,+inf)

but you see that your conclusion is wrong since the equation has only one positive root (refer to wolfram alpha)

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)

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.

Put more simply: an exponential eventually grows faster than any polynomial so 2^x will overtake x^12 at some large enough x.