Solve for x the equation 3^x+9^x=2

We'll create matching bases within equation, writting 9 as
a power of 3:

3^2 = 9

We'll
raise to x power both sides:

(3^2)^x =
9^x

We'll multiply the exponents from the
left:

3^2x = 9^x

We'll
re-write the equation:

3^x + 3^2x =
2

We'll substitute 3^x by t:

t
+ t^2 = 2

We'll re-arrange the terms and we'll subtract 2
both sides:

t^2 + t - 2 = 0

t1
= [-1+sqrt(1 + 8)]/2

t1 =
(-1+3)/2

t1 = 1

t2 =
(-1-3)/2

t2 = -4/2

t2 =
-2

We'll put 3^x = t1 => 3^x =
1

We'll create matching bases both sides, writting 1 as
3^0:

3^x = 3^0

We'll use one
to one property of exponentials:

x =
0

We'll put 3^x = t2 => 3^x = -2 impossible, since
there is no real value of x such as 3^x =
-2.

The unique solution of the given equation
is x = 0.