What is (2^{n+4}+2^{n+3})/(2^{n+3)-2^{n+2}} ?

The solution to (2^{n+4}+2^{n+3})/(2^{n+3)-2^{n+2}} is 6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (2^{n+4}+2^{n+3})/(2^{n+3)-2^{n+2}}

simplify

\frac{2 ^{n + 4} + 2 ^{n + 3}}{2 ^{n + 3} - 2 ^{n + 2}}

6

Solution steps

\frac{2 ^{n + 4} + 2 ^{n + 3}}{2 ^{n + 3} - 2 ^{n + 2}}

Factor

2^{n + 4} + 2^{n + 3} : 3 \cdot 2^{3 + n}

= \frac{3 \cdot 2 ^{3 + n}}{2 ^{n + 3} - 2 ^{n + 2}}

Factor

2^{n + 3} - 2^{n + 2} : 2^{2 + n}

= \frac{3 \cdot 2 ^{3 + n}}{2 ^{2 + n}}

Cancel

\frac{3 \cdot 2 ^{3 + n}}{2 ^{2 + n}} : 3 \cdot 2

= 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 6

\frac{\frac{3 x - 9}{20}}{\frac{2 x - 6}{5}}

- 20 ab + 11 ab

2 x^{2} + 6 x - 7 x + 8 - 3 x^{2} + 1

(7 x + 4) (7 x - 2)

What is (2^{n+4}+2^{n+3})/(2^{n+3)-2^{n+2}} ?
The solution to (2^{n+4}+2^{n+3})/(2^{n+3)-2^{n+2}} is 6