Solve x^(1/5) = floor((3x)^(1/5)).?
Not sure what this 'floor' is but i'll just use it as a variable i suppose... x^(1/5) = floor((3x)^(1/5)) =5√x (fifth root of x like a cube root..) = floor(3 X 5√x) (5√x) / 3 (5√x) = floor floor = 1/3
The only solutions are x = 0, x = 1, x = 32, x = 243, and x = 1024. --- Suppose that x is a solution to this equation. Then, since floor((3x)^(1/5)) is always an integer, then x^(1/5) must be an integer; which in turn means that x must be a 5th power. So we have ruled out all x which are not 5th powers. --- By trying small 5th powers, we see that * x = 0^5 = 0 is a solution * x = 1^5 = 1 is a solution * x = 2^5 = 32 is a solution * x = 3^5 = 243 is a solution * x = 4^5 = 1024 is a solution and that no larger 5th powers appear to work, and no negative 5th powers appear to work. Let us try to prove that the above five values are the only solutions. --- Suppose that x < 0 and x is a 5th power. Since x < 0, then we have 3x < x Since the function f(x) = x^(1/5) is increasing, then applying it to both sides of the inequality preserves the inequality, so we have (3x)^(1/5) < x^(1/5) Now, x^(1/5) is an integer, because x is a fifth power; thus, floor((3x)^(1/5)) must be a smaller integer, because (3x)^(1/5) is less than x^(1/5). So x cannot be a solution when x < 0. --- It remains to show that x cannot be a solution when x = k^5, where k is an integer greater than 4. To show this, I will first show by induction that if k is an integer greater than 4, then 3^(1/5) * k k + 1. Base case: 3^(1/5) * 5 5 + 1 by a calculator. Inductive step: Suppose that 3^(1/5) * n n + 1. Clearly also 3^(1/5) 1. Adding these two inequalities gives 3^(1/5) * n + 3^(1/5) n + 2. Factoring the left and rewriting the right gives 3^(1/5)(n + 1) (n + 1) + 1, which is what we wanted to show. --- Now, suppose that x = k^5, where k is an integer greater than 4. Then (3x)^(1/5) = 3^(1/5) * x^(1/5) = 3^(1/5) * (k^5)^(1/5) = 3^(1/5) * k, and x^(1/5) = (k^5)^(1/5) = k. But by the induction above, 3^(1/5) * k k + 1, so (3x)^(1/5) x^(1/5) + 1, which means that floor((3x)^(1/5)) x^(1/5). So x cannot be a solution to the equation.
