Exponential equation

4^x + 6^x = 9^x
(2·2)^x + (2·3)^x = (3·3)^x
Since (2·3)^x never equals to zero, we can divide both sides of this equation by it without losing any solutions.
The result is
(2/3)^x + 1 = (3/2)^x
Let y=(2/3)^x
Then y + 1 = 1/y
Since y is always positive, we can multiply both sides by it getting an equivalent equation
y² + y − 1 = 0
Solutions to this quadratic equation are
y_1,2 = (−1 ± √5)/2
Therefore, x_1,2 = log_2/3[(−1 ± √5)/2]