Logarithm Problem Involving Radicals

[Mindrust]

 

How does one go about solving this algebraically? I tried creating a new variable (let a = log(base5)x) to get a^(1/2) + a^(1/3) = 2 but not sure what to do from there... If I can figure out how to get rid of the radicals, the rest should be simple.

 

[EDIT ] A friend helped me with the answer.

 

Let y = log_5(x)Let z = y^(1/6)

 

Then the given equation becomes z^3 + z^2 = 2, that is (z-1)(z+1+i)(z+1-i) = 0. The solutions are r = 1, s = -1-i, and t=-1+i.

 

Then x = 5^(r^6) =, 5^(s^6), 5^(t^6) are solutions of the original equation. The only real valued solution is x = 5.

 

Alternatively, you can arrange the equation into cubic form, and get the roots using the cubic formula.

Edited May 1, 2011 by Mindrust