Solve:
a) 1/(2^x) = 1/(x+2)
b) 1/(2^x) > 1/(x^2)
2 answers
x> (7i + 1) / 2
for the first one, cross-multiply
2x^2 = x+2
2x^2 - x - 2 = 0
use the quadratic formula to solve.
For the second, let's see if the corresponding functions intersect, let
1/(2^x) = 1/(x^2)
2x^2 = x^2
true only for x = 0
but x cannot be zero in the original,
so they do not intersect.
clearly for │x│>1
2x^2 > x^2, so
1/(2x^2) < 1/x^2 is false
e.g.
let x=5
1/50 > 1/25 ?? ---> false
but for │x│ < 1 it is true
(the denominator of the first is bigger the second, so the first fraction is smaller)
e.g.
x = .7
1/((2)(.49)) < 1/(.49) ?? ---> true
2x^2 = x+2
2x^2 - x - 2 = 0
use the quadratic formula to solve.
For the second, let's see if the corresponding functions intersect, let
1/(2^x) = 1/(x^2)
2x^2 = x^2
true only for x = 0
but x cannot be zero in the original,
so they do not intersect.
clearly for │x│>1
2x^2 > x^2, so
1/(2x^2) < 1/x^2 is false
e.g.
let x=5
1/50 > 1/25 ?? ---> false
but for │x│ < 1 it is true
(the denominator of the first is bigger the second, so the first fraction is smaller)
e.g.
x = .7
1/((2)(.49)) < 1/(.49) ?? ---> true
1 answer