Question

16 × 2^n+1 - 4× 2^n ÷ 16 × 2n+2- 2×2^n+2 .
Simplify the following ​

Answer 1

$\underline{\underline{\pink{ \mathfrak{ Question}}}}$

• $\sf \dfrac{16 \times {2}^{n + 1} - 4 \times {2}^{n} }{16 \times {2}^{n + 2} - 2 \times {2}^{n + 2} }$

$\underline{\underline{\red{ \mathfrak{ Answer \: = \blue{\sf \dfrac{1}{2}} }}}}$

$\underline{\underline{\orange{ \mathfrak{ Step \: by \: step \: explanation }}}}$

$= \sf \dfrac{16 \times {2}^{n + 1} - 4 \times {2}^{n} }{16 \times {2}^{n + 2} - 2 \times {2}^{n + 2} }$

$= \sf \dfrac{ {2}^{4} \times {2}^{n + 1} - {2}^{2} \times {2}^{n} }{ {2}^{4} \times {2}^{n + 2} - 2 \times {2}^{n + 2} }$

$= \sf \dfrac{ {2}^{n + 5} - {2}^{n + 2} }{ {2}^{n + 6} - {2}^{n + 3} }$

$= \sf \dfrac{ {2}^{n + 5} - {2}^{n + 2} }{2 \times {2}^{n + 5} - 2 \times {2}^{n + 2} }$

$= \sf \dfrac{ \: \: {2}^{n + 5} - {2}^{n + 2} }{2( {2}^{n + 5} - {2}^{n + 2} ) }$

$= \sf \dfrac{ \cancel{ {2}^{n + 5} - {2}^{n + 2 } } ^{} }{2 \cancel{ ( {2}^{n + 5} - {2}^{n + 2} ) } ^{} }$

$\sf \: = \green{\dfrac{1}{2} }$

Laws of exponents

• $\tt{a}^{m} \times {a}^{n} = {a}^{m + n} \: \: \: \: \: \: \: \: \: \: \sf {a}^{m} \div {a}^{n} = {a}^{m - n} \\ \sf{( {a}^{m} ) ^{n} = {a}^{mn} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a {}^{m} \times {n}^{m} = (ab) ^{m} } \\ \sf{a}^{0} = 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\frac{ {a}^{m} }{ {b}^{m} }= \left( \frac{a}{b} \right) ^{m} }$

Answer 2

Step-by-step explanation:

hope this helps you

and you get your answer