kinda weird, but if you've written it correctly,
log(4/x) (x^2 - 6) = 2
raise 4/x to both sides and you have
x^2-6 = (4/x)^2
x^2 - 6 = 16/x^2
x^4 - 6x^2 - 16 = 0
(x^2+2)(x^2-8) = 0
x = ±√2 i or ±2√2
(x+√(x^2-5))/5 =
(x+√(x^2-5))(x-√(x^2-5))
---------------------
5(x-√(x^2-5))
x^2 - (x^2-5)
----------------
5(x-√(x^2-5))
1/(x-√(x^2-5))
So, log of one is -log of its reciprocal.
1. Solve log(4/x) (x^2 - 6) = 2. (note: the (4/x) is underscored)
2. Show that loga [(x + {sqrtx^2-5}) / 5] = -loga (x - (sqrtx^2-5))
1 answer