The number of real roots of the equation e^6x-e^4x-2e^3x-12e^2x+e^x+1=0 is

Question

The number of real roots of the equation $e^{6x}-e^{4x}-2e^{3x}-12e^{2x}+e^x+1=0$ is

Accepted Answer

Grouping: $(e^{3x}-1)^2-e^x(e^{3x}-1)=12e^{2x} \implies (e^{3x}-1)(e^{3x}-e^x-1)=12e^{2x}.$

Analysis of this equation (numerically/graphically) shows it has exactly 2 real solutions.

Solutions · 1