That's not the only solution, though. For x = 1, LHS is bigger. For x = 2, RHS is bigger. Both sides are continuous, so there must be another solution between 1 and 2, but very close to 1. I don't know how to solve it, but WolframAlpha says it's about 1.0025851, and my calculator agrees. As far as how to find the integer solution, the way I did it is to take the base 5 log, you get x = 625 log_5(x) Let y = log_5(x) 5^y = 5^4 y 5^(y-4) = y If y is an integer, y must be a power of 5. Try y=5, you get 5^(5-4) = 5, which is true x = 5^y = 5^5 = 3125
Niceeee.
That's not the only solution, though.
For x = 1, LHS is bigger.
For x = 2, RHS is bigger.
Both sides are continuous, so there must be another solution between 1 and 2, but very close to 1. I don't know how to solve it, but WolframAlpha says it's about 1.0025851, and my calculator agrees.
As far as how to find the integer solution, the way I did it is to take the base 5 log, you get
x = 625 log_5(x)
Let y = log_5(x)
5^y = 5^4 y
5^(y-4) = y
If y is an integer, y must be a power of 5. Try y=5, you get
5^(5-4) = 5, which is true
x = 5^y = 5^5 = 3125
Thanks for your feedback. Indeed the problem can also be solved using logs!