Solve for x:
6^(1+x)+6^(1-x)=37
6^(1+x)+6^(1-x)=37
6*6^(x) + 6*6^(-x) = 37
6^(x) + 6^(-x) = 37/6
6^x + 1/6^(x) = 37/6
6^x*6^x + 1 = (37/6)*6^x
6^(2x) - (37/6)*6^(x) + 1 = 0
[6^(x)]^2 - (37/6)*[6^(x)] + 1 = 0
Let y = 6^x
y^2 - (37/6)y + 1 = 0
Factor, complete the square, use the quadratic formula, whatever. You get:
y = 1/6 or y = 6.
Thus
6^x = 1/6 ==> x = -1
or
6^x = 6 ==> x = 1
Thus x = -1 or x = 1.
-Dan