The value of $\frac{2^{4}-2}{2^{3}-1}$ is

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-020.jpg?height=262&width=355&top_left_y=1582&top_left_x=1313)
(A) 1
(B) 0
(C) $\frac{7}{4}$
(D) $\frac{4}{3}$
(E) 2
We note that $2^{3}=2 \times 2 \times 2=8$ and $2^{4}=2^{3} \times 2=16$.

Therefore, $\frac{2^{4}-2}{2^{3}-1}=\frac{16-2}{8-1}=\frac{14}{7}=2$.

Alternatively, $\frac{2^{4}-2}{2^{3}-1}=\frac{2\left(2^{3}-1\right)}{2^{3}-1}=2$.

ANSWER: (E)