The number of real roots of equation $\log_{e}x + ex = 0$
- $0$ (zero)
- $1$
- $2$
- $3$
Let f(x) = log(x) + ex. The derivative f'(x) = 1/x + e. Since x must be positive for log(x) to be defined, f'(x) is always positive for x > 0. Thus, the function is strictly increasing. As x approaches 0, f(x) approaches -infinity, and as x approaches infinity, f(x) approaches infinity. By the intermediate value theorem, there is exactly one root.
The equation log_e(x) + ex = 0 can be rewritten using logarithmic properties as log_e(x) = -ex, which implies x = e raised to the power of -ex. By substituting y = ex, this transforms into y(e^-y) = 1, or equivalently ye^y = 1. The function g(y) = ye^y is negative for y < 0, zero at y = 0, and strictly increasing to infinity for y > 0. Because g(y) equals 1 at exactly one point in the positive domain, the equation has exactly one real root.