How many intersections do 2^x and x^12 have in the x-y plane?

[FeDeX_LaTeX]

Hello,

I found a question in a practice GRE Mathematics test, but I wasn't sure if I was using the right method to solve it. The question is to find the number of intersections in the x-y plane of the functions 2^x and x¹². So I got;

\frac{\ln{2}}{12} = \frac{\ln{x}}{x}

I don't think a calculator is allowed for this paper, but I was wondering if this was the right method. I was going to rewrite ln(x) as a Taylor series approximation, divide the whole thing by x and then see where that got me. Is this the wrong way to do it?

Other than that I'm not sure how to do it. Given that it is in an exam it can't require a computer program to solve it either... I would appreciate some help on this. Plus, I don't know if the graph of 2^x may 'overtake' the graph of x^12, or not... so, not sure what to do. Help would be appreciated.

The possible answers are;

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

I'm pretty sure A is wrong, but I don't know about the rest. If I had to guess I would say at least 2.

You also only get roughly 2.5 minutes per question in this test (66 questions in 170 minutes) so I'm not sure if this method is even practical... is there something obvious I'm missing? I know the general shapes of both graphs, so if one were to overtake another, the derivative of 2^x would equal the derivative of x^12 at some point and then it would be less than the derivative of x^12 after that, right? But even if that were true that would just give me one root.

EDIT: Also, this is not a homework question, just bored in my summer holidays.

I thought about the graphs of Kx and ln(x) and I'm more convinced there is 1 intersection pretty early on (at x = 0, Kx is at origin, ln(x) is rising very fast). I think the answer is 2 or 3.

 

[atomthick]

Study the function 2^x - x^12 look for changes in sign for its values.

 

[uart]

This question may just be testing basic curve sketching and knowledge of properties of elementary functions.

If you make even a rough sketch (of x^12 and 2^x) then you'll find two points of intersection (near the origin) very quickly. To know that there is a third point of intersection for larger values of "x" you only need to know about the limiting behavior of exponentials versus powers.

 

[FeDeX_LaTeX]

The answer is 3, then? There is definitely an intersection on the LHS of the y-axis and one close to the origin on the RHS of the y-axis.

Does a^x approach infinity faster than x^a (where a is a constant)? I took the derivative of both 2^x and x^12 and I think the exponential would approach infinity faster.

 

[uart]

The answer is 3, then? There is definitely an intersection on the LHS of the y-axis and one close to the origin on the RHS of the y-axis.

Does a^x approach infinity faster than x^a (where a is a constant)? I took the derivative of both 2^x and x^12 and I think the exponential would approach infinity faster.

Yes, and I think that this was the piece of knowledge that this question was really aiming to test.

You can show that for any a>1 and for any finite n, that a^x > x^n for large enough x. So knowing this property of the exponential is sufficient to prove that there is a third solution.

BTW. In a high school test you probably wouldn't need to prove that property to use it. Simply stating it (and it being correct and a well know property) should be enough.

 

[FeDeX_LaTeX]

Hi,

Could you show me the proof? I can't find it via Google... thanks. The closest thing I found was an explanation on PurpleMath but there was no proof. I also saw an evaluation of the limit of an exponential function (t approaching infinity) on the exponential function wiki article, but I don't think that is enough to prove it.

 

[uart]

Hi, Could you show me the proof?

Yeah sure. You were already half way there in your very first post when you deduced:\frac{\ln{2}}{12} = \frac{\ln{x}}{x}.

Lets just generalize that by replacing "2" with "a" (a>1) and replacing "12" with "n", where n is an arbitrarily large positive number. Proceeding as you did in the first post you find that for a^x > x^n you require:

\frac{\ln{x}}{x} < \frac{\ln{a}}{n}.

The RHS in the above inequation is a positive number, so it's sufficient to prove that:

\stackrel{\lim}{x \rightarrow \infty} \left\{ \frac{\ln{x}}{x} \right\} = 0

With one application of L'Hopitals rule (differentiate top and bottom) this is equivalent to

\stackrel{\lim}{x \rightarrow \infty} \left\{ \frac{1}{x} \right\} = 0

 

[FeDeX_LaTeX]

Thanks! I didn't think it would be that simple. I completely forgot about using L'Hopitals rule for a divergent limit.