Suppose 64x−14x−1=2562x\frac{64^{x-1}}{4^{x-1}} = 256^{2x}4x−164x−1=2562x Then, what is x? −23\dfrac{-2}{3}3−2 −13\dfrac{-1}{3}3−1 000 14\dfrac{1}{4}41 38\dfrac{3}{8}83 Solution Given 64x−14x−1=2562x\frac{64^{x-1}}{4^{x-1}} = 256^{2x}4x−164x−1=2562x Then, since 64=4364 = 4^364=43 and 256=44256 = 4^4256=44, 64x−1=2562x⋅4x−143(x−1)=48x+(x−1)43x−3=49x−13x−3=9x−1−6x=2x=−13\begin{aligned} 64^{x-1} &= 256^{2x} \cdot 4^{x - 1} \\ 4^{3(x - 1)} &= 4^{8x + (x - 1)} \\ 4^{3x - 3} &= 4^{9x - 1} \\ 3x - 3 &= 9x - 1 \\ -6x &= 2 \\ x &= \frac{-1}{3} \end{aligned}64x−143(x−1)43x−33x−3−6xx=2562x⋅4x−1=48x+(x−1)=49x−1=9x−1=2=3−1 The answer is b.