Let S be the set of all real roots of the equation 3^x(3^x-1)+2 = |3^x-1|+|3^x-2|.

Question

Let S be the set of all real roots of the equation $3^x(3^x-1)+2 = |3^x-1|+|3^x-2|$. Then S:

Accepted Answer

Let $y=3^x$ ($y>0$), so $y(y-1)+2 = |y-1|+|y-2|$. Case $y>2$: $y^2-y+2=(y-1)+(y-2)=2y-3 \implies y^2-3y+5=0$, $D=9-20<0$ --- no solution. Case $1\le y\le2$: $y^2-y+2=(y-1)+(2-y)=1 \implies y^2-y+1=0$, $D=1-4<0$ --- no solution. Case $y\le1$: $y^2-y+2=(1-y)+(2-y)=3-2y \implies y^2+y-1=0 \implies y=\dfrac{-1\pm\sqrt5}{2}$. Since $y=3^x>0$ and $y\le1$: only $y=\dfrac{\sqrt5-1}{2}\approx0.618$ is valid. So there is exactly one real solution, $x=\log_3\!\left(\dfrac{\sqrt5-1}{2}\right)$ --- S is a singleton.

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