clearly x must be positive for logx to exist.
Now, I'm not sure what tools you have at your command for working on this.
You could define f(x) = 3x^2-2x^3-log((x^2+1)/x) and find its zeros numerically.
Or, if you have calculus available,
3x^2-2x^3 has a single local maximum at (1,1).
Now, if we define
f(x) = 3x^2-2x^3
g(x) = log (x^2+1)/x
f(0) < g(0)
f(1) > g(1)
f(2) < g(2)
So, the graphs intersect on the intervals (0,1) and (1,2)
That is, there are 2 solutions. You can see them at
http://www.wolframalpha.com/input/?i=3x%C2%B2-2x%C2%B3%3Dlog(x%C2%B2%2B1)-logx
Number of solutions satisfying the equation, 3x²-2x³=log(x²+1)-logx is
a) 1
b) 2
c) 3
d) none
1 answer