Question

if x+y=0 , 2^x+2^y=40 then find x,y​

Answer 1

Answer:

x = y

=>

$2 {x}^{2} + {2x}^{2} = 40 \\ {4x}^{2} = 40 \\ {x}^{2} = 40 \div 4 \\ {x}^{2} = 10 \\ x = \sqrt{10}$

Answer 2

-x = y
2^x+2^-x=40
2^x+1/2^x=40
Let 2^x be t
t + 1/t =40
Taking LCM,
t^2+1 =40
t^2=39
+- √39=t
Now, substitute t