Answer
$(3,1)$
$(1,3)$
Work Step by Step
Given the system of equations:
1. $2^x + 2^y = 10$
2. $4^x + 4^y = 68$
Reformat equation 2
1. $2^x + 2^y = 10$
2. $(2^x)^2 + (2^y)^2 = 68$
The question asks to find all real solutions.
Substitute $2^x = 10 - 2^y$
$(10-2^y)^2 + (2^y)^2 = 68$
$100 - 20 * 2^y + (2^y)^2 + (2^y)^2 = 68$
$2(2^y)^2 - 20 * 2^y + 32 = 0$
$2((2^y)^2 - 10 * 2^7 + 16) = 0$
$2((2^y - 2)(2^y - 8) = 0 $
$2^y = 2$
$y = 1$
$2^y = 8$
$y = 3$
Substitute the values into the equation
$2^x = 10 - 2^1$
$2^x = 8$
$x = 3$
$2^x = 10 - 2^3$
$2^x = 2$
$x = 1$
All Solutions:
$(3,1)$
$(1,3)$
