2100.52 – Rational Expressions, Laws of Exponents

Tags:Problem Set 10GeometrySymbol-Pushing

Suppose

64x−14x−1=2562x\frac{64^{x-1}}{4^{x-1}} = 256^{2x}

Then, what is x?

  1. −23\dfrac{-2}{3}
  2. −13\dfrac{-1}{3}
  3. 00
  4. 14\dfrac{1}{4}
  5. 38\dfrac{3}{8}
Solution

Given

64x−14x−1=2562x\frac{64^{x-1}}{4^{x-1}} = 256^{2x}

Then, since 64=4364 = 4^3 and 256=44256 = 4^4,

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}

The answer is b.