Is the answer option 3 ?
I'm getting the only solution as ln2 .
The number of solution of the equation e2x + ex + e-2x + e-x = 3(e-2x+ ex) is
1. 0
2. 2
3. 1
4. more than 2
If you let y=ex, then the equation is f(y) = y^3 + 2y^2+1-\frac{2}{y} = 0
We note that f'(y)>0. Hence it has at most one solution.
As y \rightarrow 0, f(y) \rightarrow -\infty and as y \rightarrow \infty, f(y) \rightarrow \infty
So there exists a positive root to the given equation ( we need +ve as y=ex>0)
Hence the given equation has exactly one root