Find the value of x in (4^5⋅x+32^2)⋅2^5=2^16⋅x

Find the value of x in (4⁵⋅x+32²)⋅2⁵=2¹⁶⋅x

First we transform everything into powers of 2 as all the numbers involved in this equation are multiples of 2.

||    4⁵ = (2²)⁵ = 2^2*5= 2¹⁰

||    32²  = (2⁵)² = 2^2*5= 2¹⁰
dividing over 2⁵ on both sides and putting the powers of 2 we get
2¹⁰ x + 2¹⁰ = 2¹¹x ------ 2¹⁶/2⁵ = 2^16-5 = 2¹¹

we factorise on the left hand side of the equation

2¹⁰(x+1) = 2¹¹ x --------- we divide over 2¹⁰

x+1 = 2x -------- 2¹¹/2¹⁰ = 2^11-10 = 2

Take away x on both sides -----> x =1