Question

please give the answer

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{1}{16} \times {( \frac{1}{2} )}^{2} = {( \frac{1}{2} )}^{3(x - 2)}$
We can also write as ,

$\frac{ {1}^{4} }{ {2}^{4} } \times {( \frac{1}{2} )}^{2} = {( \frac{1}{2} )}^{3(x - 2)} \\ \\ {( \frac{1}{2} )}^{4} \times {( \frac{1}{2} )}^{2} = {( \frac{1}{2} )}^{3 \times x - 3 \times 2} \\ \\ {( \frac{1}{2} )}^{4} \times {( \frac{1}{2} )}^{2} = {( \frac{1}{2} )}^{3x - 6}$

Using the identity :

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

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

As the bases are same, so putting the powers equal we get :

$6 = 3x - 6 \\ \\ 6 + 6 = 3x \\ \\ 12 = 3x \\ \\ x = \frac{12}{3} \\ \\ x = 4$


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...
☺☺

Answer 2

$Given : \frac{1}{16} * ( \frac{1}{2} )^2 = ( \frac{1}{2}) ^{3(x-2)}$

$\frac{1}{16} * \frac{1}{4} = ( \frac{1}{2}) ^{3(x-2)}$

$\frac{1}{64} = ( \frac{1}{2} )^{3(x-2)}$

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

6 = 3(x-2)

6 = 3x - 6

3x = 12

x = 4.


Hope this helps!