Question

find the value of x such that:♤ see attachment​

Answer 1

Answer:

gyan prakash verma

Step-by-step explanation:

$( \frac{p}{q} )^{3x+ 2} = ( \frac{q}{p}) ^{2 - x}$

$= ( \frac{q}{p}) ^{2 - x} \times \frac{q}{p}^{3x + 2}$

$so \\ according \: to \: law \: = {a}^{t} \times {a}^{u} = {a}^{t} \times ^{u}$

$= (\frac{q}{p} ) ^{2 - x} + ^{3x + 2}$

$3x + 2 = 2 - x \\ 3x + x = 2 - 2 \\ 4x = 0 \\ x = 0$