If-x-and-y-are-positive-integers-and-2x-4y-3-then-find-16-x-256-y-

Question Number 179650 by Linton last updated on 31/Oct/22

I f x a n d y a r e p o s i t i v e

i n t e g e r s a n d 2 x - 4 y = 3

t h e n f i n d \frac{16^{x}}{256^{y}}

Commented by Rasheed.Sindhi last updated on 31/Oct/22

\underset{e v e n}{\underset{⏟}{2 x}} - \underset{e v e n}{\underset{⏟}{4 y}} = \underset{o d d}{\underset{⏟}{3}}

D i f f e r e n c e o f t w o e v e n n u m b e r s

{c a n}^{'} t b e o d d .

2 x - 4 y = 3 \Rightarrow x - 2 y = \frac{3}{2}

l e f t s i d e i s a n i n t e g e r w h i l e r i g h t

s i d e i s n o n - i n t e g e r r a t i o n a l!

C o n t r a d i c t i o n .

Commented by CElcedricjunior last updated on 31/Oct/22

2 x - 4 y = 3 => x = 2 y + \frac{3}{2}

o n a \frac{16^{x}}{256^{y}} = \frac{16^{2 y + \frac{3}{2}}}{16^{2 y}} = 16^{2 y - 2 y + \frac{3}{2}} = 16^{\frac{3}{2}} = 4^{3}

=> \frac{16^{x}}{256^{y}} = 64

\dots \dots \dots \dots \dots . l e c e l e b r e c e d r i c j u n i o r \dots \dots

Commented by Rasheed.Sindhi last updated on 31/Oct/22

\frac{16^{x}}{256^{y}} = \frac{4^{2 x}}{4^{4 y}} = 4^{2 x - 4 y} = 4^{3} = 64

[B u t x & y {c a n}^{'} t b e p o s i t i v e i n t e g e r s]