Question

plz answer
And plzzz no abusive replies
If u don't want to ans then ignore but don't abuse

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{16 \times {2}^{x + 1} - 4 \times {2}^{x} }{16 \times {2}^{x + 2} - 2 \times {2}^{x + 2} }$

We can also write it as

$\frac{ {2}^{4} \times {2}^{x + 1} - {2}^{2} \times {2}^{x} }{ {2}^{4} \times {2}^{x + 2} - 2 \times {2}^{x + 2} }$

Using the identity :

${a}^{m} \times {a}^{n} = {a}^{m + n}$

$= \frac{ {2}^{4 + x + 1} - {2}^{2 + x} }{ {2}^{4 + x + 2} - {2}^{1 + x + 2} } \\ \\ = \frac{ {2}^{5 + x} - {2}^{2 + x} }{ {2}^{6 + x} - {2}^{3 + x} }$

Using the identity :

$\frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$= {2}^{(5 + x) - (6 + x)} - {2}^{(2 + x) - (3 + x)} \\ \\ = {2}^{5 + x - 6 - x} - {2}^{2 + x - 3 - x} \\ \\ = {2}^{ - 1} - {2}^{ - 1} \\ \\ = \frac{1}{2} - \frac{1}{2} \\ \\ = 0$


Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
☺☺