I used Python code

for a in xrange(10):
    for b in xrange(10):
        for c in xrange(10):
            if((a**b)*(c**a) == (1000*a+100*b+10*c+a)):
                print a, b, c, a

Output

2 5 9 2

Hi everyone. Firstly, the equation is a^b * c^a = abca. The method I used is an assumption. The range for all of the 4 digits is 1-9. So I made the assumption from the smallest digit.

Let a = 2.

So , the equation turns into

2^b * c^2 = 2bc2

From the equation, the answer is a four-digit-number.

Another assumption is made.

Let c = 9. The only reason to this is that putting the largest range of the number will decrease the time taken.

So, 2^b * 9^2 = 2bc2.

9^2 = 81.

The answer must be a four digits containing 2 in the front and another 2 at the back.

Let b = 9.

2^9 = 2080. The multiplication of 2080 and 81 has exceeded the range for the answer which is a four-digits-number containing 2 in the front and 2 at the back.

The following assumptions are made by decreasing the power of 2. Until the the following assumption, 2^5.

2^5 = 32.

32*81 = 2592.

The answer 2592 is correct as the arrangement of the combination of numbers, which is

a^b*c^a=abca

2^5*9^2=2592

a=2

b=5

c=9

This is the method I used to solve this problem. If anyone finds this wrong, please give me some feedback for me to improve myself.

The method I used is assumptions. I hope all of you who can solve the problem with other methods, do write your solutions here for me and the others to learn for more.

My name is TAN JIA JIN, and I come from Malaysia. Thank you. Have a nice day.

Sincerely, TAN.

Python:

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def goal(n):
    a,b,c,d = (int(i) for i in str(n))
    return n == a**b*c**d
for n in xrange(1000,10000):
    if goal(n):
        print "Answer:", n
        break

I'm too lazy to write codes, and i like "drag and drop" so... :P

0.04 second , fast enough for me. :P

Java solution -

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public static void main(String[] args){
    final int min = 1000, max = 9999;
    for(int i = min; i <= max; i++) 
        if(i == div(i)){
            System.out.println(i);
            return;
        }
 }

private static double div(int z){
    String t = String.valueOf(z);
    double[] val = new double[4];
    for (int i = 0; i < 4; i++)
        val[i] = t.charAt(i) - '0';
    return Math.pow(val[2], val[0]) * Math.pow(val[0], val[1]);
} 

Execution time ~0.002 second

This number is a Friedman Number

The easiest approach would be to assume values of a and try for $b = c = 9 ~ and~ see~ if~~ a^9 * 9^a$ is a four digit number. If not got to next value of a. For a =0, 1, we do not get a four digit number.
Next $~~a = 2.~~~~ 2^9*9^2=41472 ~~~$ We get a number greater than four digits. $b=8,~~ ~2^8*9^2=20736..NO…~~~~~~~~~~~~ b=7,~~~ 2^7*9^2=10368..NO..\\ b=6, ~~~6^8*9^2=5184..~~~~it~ is~ four~ digits,~ but~ not~ as~we~desire.\\ b=5, 2^5*9^2=2592…..~~~~\boxed{ 2592 }$