If $$3x - y = 5$$, find the value of $$\frac{8^x}{2^y}$$.

Given that

$$\frac{8^x}{2^y}$$.            equation 1

$$3x - y = 5$$                      equation 2

Now find the value of x

$$3x  = 5+y$$

$$x=\frac{5+y}{3}$$

Now put the value of x in equation  1

we get

= $$\frac{8^\frac{5+y}{3}}{2^y}$$. 

we know that $$2^3=8 $$             equation 3

Using the above equation 3

= $$\frac{2^{3(5+y)/3}}{2^y}$$.

= $$\frac{2^{(5+y)}}{2^y}$$.

Now minimize the equation

= $${2^{(5+y-y)}}$$.

= $${2^{5}}$$.

= 32           Ans