Suppose64x−14x−1=2562x\dfrac{64^{x-1}}{4^{x-1}} = 256^{2x}4x−164x−1=2562xThen, what is x?−23\dfrac{-2}{3}3−2−13\dfrac{-1}{3}3−100014\dfrac{1}{4}4138\dfrac{3}{8}83 Solution Given64x−14x−1=2562x\dfrac{64^{x-1}}{4^{x-1}} = 256^{2x}4x−164x−1=2562xThen, 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 &= \dfrac{-1}{3} \end{aligned}64x−143(x−1)43x−33x−3−6xx=2562x⋅4x−1=48x+(x−1)=49x−1=9x−1=2=3−1The answer is b.