Question

slove this problem please please please please please please please

Answer 1

Solution :-

◾Given :-

$\left( \dfrac{-3}{2}\right)^6 \times \left(\dfrac{4}{9}\right)^3 = \left(\dfrac{1}{2}\right)^{3x}$

◾Now by simplification of above :-

$\implies \left( \dfrac{-3}{2}\right)^6 \times \left(\dfrac{2^2}{3^2}\right)^3 = \left(\dfrac{1}{2}\right)^{3x}$

$\implies \left( \dfrac{-3}{2}\right)^6 \times \left(\left(\dfrac{2}{3}\right)^2\right)^3 = \left(\dfrac{1}{2}\right)^{3x}$

$\implies \left( \dfrac{-3}{2}\right)^6 \times \left(\dfrac{2}{3}\right)^6 = \left(\dfrac{1}{2}\right)^{3x}$

$\implies \left( \dfrac{9^3}{2^6}\right) \times \left(\dfrac{2^6}{9^3}\right) = \left(\dfrac{1}{2}\right)^{3x}$

$\implies \left( \dfrac{\cancel{9^3}}{\cancel{2^6}}\right) \times \left(\dfrac{\cancel{2^6}}{\cancel{9^3}}\right) = \left(\dfrac{1}{2}\right)^{3x}$

$\implies 1 = \left(\dfrac{1}{2}\right)^{3x}$

◾Now we will consider $\left(\frac{1}{2}\right)^{0}$ = 1 .

▪️Then :-

$\implies \left(\dfrac{1}{2}\right)^{0} = \left(\dfrac{1}{2}\right)^{3x}$

◾As bases are same :-

$\implies 0 = 3x$

$\implies x = 0$

So value of x = 0