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 = 22*5= 210
|| 322 = (25)2 = 22*5= 210
dividing over 25 on both sides and putting the powers of 2 we get
210 x + 210 = 211x ------ 216/25 = 216-5 = 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 = 211-10 = 2
Take away x on both sides -----> x =1
LC
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