6.1. (New York, 78). Solve the equation $8^{x}(3 x+1)=4$.

Solve the equation $8^{x}(3 x+1)=4$.
6.1. The number $x=1 / 3$ is a solution to the given equation. Let's prove that there are no other solutions. For $x>-1 / 3$, the functions $y_{1}(x)=8^{x}$ and $y_{2}(x)=3 x+1$ take positive values and increase, so their product (the left side of the equation) is also an increasing function. Therefore, on the interval $(-1 / 3 ;+\infty)$, the equation cannot have more than one solution. Further, for $x \leqslant-1 / 3$, we have $y_{1}(x)>0, y_{2}(x) \leqslant 0$, which means $y_{1}(x) y_{2}(x) \leqslant 0$. Therefore, on the interval $(-\infty ;-1 / 3]$, the equation has no solutions. Thus, we get the answer: $1 / 3$.