Question 560634
x = 0 works, but we need to prove it is the only solution. Note that we can write the equation as


*[tex \LARGE 9^x + 6^x = 2(4^x)]


Dividing both sides by 4^x,


*[tex \LARGE (\frac{9}{4})^x + (\frac{6}{4})^x = 2]


As you can see, the LHS is a strictly increasing function of x (bijective or one-to-one), so there can only be one value of x in which the LHS equals 2. We have already found it to be x=0.