Is There Exponent...?

Question

That if u reverse the base with the exponent it will equal the same. one besides 2 to the 4th power and 4 to the 2nd power.

Answer 1

There are many. For instance, 2^x=x^2 has three solutions: x=2, x=4, and x≈-0.76666469596212306 (this solution is irrational and has no closed-form expression). 4^x=x^4 also has these same three solutions.
6^x=x^6 has three solutions, the obvious one at x=6, one at x≈1.6242438458589112 and a second at x≈-0.78987685692164455 (again, these solutions are irrational and have no closed-form expression).

In general, for any number n > e, n^x=x^n will have at least one solution in the range (1, n). The reason is as follows: Let f(x) = n^x-x^n. Then f'(x)=n^x ln n - nx^(n-1). At x=n, f'(x)=n^n (ln n - 1), which is positive for n>e. Now, f(x)=0 at x=n, and is increasing at the same point, therefore for any ε there exists a positive δ<ε such that f(n-δ) is strictly negative (intuitively, this means that since the function is increasing as it hits zero, then just before it hit zero it must have been negative), and therefore some point c=n-δ strictly between 1 and n where f(c) is negative. However, f(1) is clearly positive. Therefore, by the intermediate value theorem, there exists a zero of f(x) at some point in the interval (1, c), which is a subinterval of (1, n). Thus, there exists a point in the open interval (1, n) such that n^x-x^n=0 and therefore a point where n^x=x^n. Thus, every problem of the form n^x=x^n provably has at least 1 nontrivial solution (the trivial solution being x=n) provided n>e.

If 1n+δ>n where f(x)=0, and thus a nontrivial solution greater than n.

Also we can show that if n is a positive even integer, there is always another nontrivial solution in the range (-1, 0). Again, define f(x)=n^x-x^n. At x=-1, f(x)=1/n-1, which is clearly negative. But at x=0, f(x)=1, which is positive. By the intermediate value theorem, there must be a point in the open interval (-1, 0) such that f(x)=0 and x^n=n^x.

Edit: corrected typo.

Answer 2

Generally it's not true unless the base and the exponent are the same number. The 2^4 and 4^2 would be the exeception.

Answer 3

Usually only if the base and exponent are the same, i.e. 1^1 2^2, etc. or 2^4 and 4^2 . Otherwise, no

Answer 4

i was gonna say when the exponent equals the base, then it is. 3^3=3^3. but i dont know.

Answer 5

Generally, it's not true. It is true that 2^4 = 4^2.

Answer 6

You are asking for points (x,y) where y^x = x^y. Besides the obvious solution, y = x, the only integer solutions are (2,4) and (4,2).
There are non-integer solutions, plot it and see.
You can try this for yourself if you have access to Matlab or similar. Define two functions such as
Z = x^y.
and
W = y^x.
Then plot the surfaces and see where the intersect.

Answer 7

It's possible.What do you think?