An algebra problem by Mohammad Al Ali

$3^{2(x-1)} - 3^{x-1} = 2$

Let $y = 3^{x-1}$

$y^{2} - y - 2 = 0$

So $y = 2, -1$
As $y = 3^{x-1}$ , so $y$ can only be positive i.e. $2$ .

So we have -
$3^{x-1} = 2$
$(x-1) \log_{10} 3 = \log_{10} 2$
$(x-1) = \log_{3} 2$
$x = \log_{3} 2 + \log_{3} 3$
$x = \log_{3} 6$

First, we write the equation using a common base for the exponential expressions: $(3^{2})^{x-1} - 3^{x-1} - 2 = 3^{2x-2} -3^{x-1} -2 = 0$ . Thus

$3^{2x-2} - 3^{x-1} -2 = 3^{2x} 3^{-2} - 3^{x} 3^{-1} -2 = \frac{ 3^{2x} }{9} - \frac{ 3^{x} }{3} - 2 = 0$

Making the substitution $y = 3^{x}$ , so $y^{2} = 3^{2x}$ , we have $\frac{ y^{2}}{9} - \frac{y}{3} - 2 = 0$ . Multiplying by 9 gives $y^{2} - 3y - 18 = (y-6)(y+3)$ . Thus $3^{x} = -3, 6$ . The first has no real solution, but the second has solution $x = log_3{6}.$