How do you solve #log_(1/2) (x^3 + x) + log_(1/2) (x^4 - 2x) = 1#?

1 Answer
Apr 22, 2016

Answer:

Roots of #2x^7+2x^5-4x^4-4x^2-1=0#, with #x > 2^(1/3#. This has only one real root in (1.29, 1.3). By bisection method , we can get closer to the root. Newton-Raphson method is faster,

Explanation:

Using #log a + log b = log ab and log_a a = 1#,.

#log_(1/2)(x^3+x)(x^4-2x))=1#.

So. #(x^3+x)(x^4-2x)=1/2#
#f(x)=2x^7+2x^5-4x^4-4x^2-1=0#

It is easy to prove that this has only one real root that is positive. Also, using sign test, f(1.29) < 0 and f(1.3) > 0. The root is in (129, 1.3.
Really, rounded to 3-sd. it is 1.29.