Possible number of order pairs satisfy 16^(x^2 +y) +16^(y^2 +x) =1 is


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 171226 by SLVR last updated on 10/Jun/22

$P o s s i b l e n u m b e r o f o r d e r p a i r s$ $s a t i s f y 16^{x^{2} + y} + 16^{y^{2} + x} = 1 i s$
Commented by SLVR last updated on 11/Jun/22

$c a n a n y o n e m e p l e a s e$
Commented by SLVR last updated on 10/Jun/22

$i m e a n p a i r s (x, y) i s h o w m a n y ?$ $k i n d l y h e l p m e o u t$
Commented by mr W last updated on 11/Jun/22

$2^{4 (x^{2} + y)} + 2^{4 (y^{2} + x)} = 1$ $4 (x^{2} + y) = 4 (y^{2} + x) = - 1$ $\Rightarrow x = y = - \frac{1}{2}$
Commented by SLVR last updated on 11/Jun/22

$S o k i n d o f y o u p r o f e s s o r .$ $G r e a t l y t h a n k y o u$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum