log equation

Draw 2 graphs
y = 2 log x .... (a)
y = x log 2 .... (b)
Both of this 2 graphs are : when x is increased, y is also increased.
When x = 1, y(a) = 0, y(b) = 0.301, so y(a) < y(b) ;
when x = 3, y(a) = 0.9542, y(b) = 0.9031, so y(a) > y(b) ;
when x = 5, y(a) = 1.3979, y(b) = 1.5051, again y(a) < y(b).
From the above result, these two graph should have 2 intersection points,
one is between 1 and 3, another is between 3 and 5.
By bisection method, you can find that the result is x = 2 and x = 4.

For the equation x^2 = 2^x, also draw 2 graphs,
y = x^2 .... (c)
y = 2^x .... (d)
Graph (c) is a parabola, min pt is (0, 0),
when x = 0, y(c) < y(d), so they also have 2 intersection points.
When x = -1, y(c) = 1, y(d) = 0.5, y(c) > y(d) ;
when x = 3, y(c) = 9, y(d) = 8, y(c) > y(d).
So one of the intersection points is between -1 and 0, and the other is between 0 and 3.
By bisection method, you can find that the result is x = -0.766.. and x = 4

2012-09-17 12:07:12 補充：
Sorry, the last statement should be :
By bisection method, you can find that the result is x = -0.766... and x = 2.

2012-09-19 10:49:55 補充：
Modify :
Graph (c) is a parabola, min pt is (0, 0), graph (d) is an increasing curve,
when x = -1, y(c) = 1, y(d) = 0.5, y(c) > y(d)
when x = 0, y(c) = 0, y(d) = 1 y(c) < y(d)
when x = 3, y(c) = 9, y(d) = 8, y(c) > y(d)
when x = 5, y(c) = 25, y(d) = 32, y(c) < y(d)

2012-09-19 10:51:37 補充：
So there are 3 intersection points, one between -1 and 0, one between 0 and 3,
and one between 3 and 5.
By bisection method, the result is x = -0.766..., x = 2 and x = 4.

There's another solution other than 2.

2log x = x log 2

log x^2 = log 2^x

x^2 = 2^x

x = 2

2012-09-16 18:44:25 補充：
x = 4 is another solution