[1/2]^x + 1/3]^x+3 = 9

To solve the equation [1/2]^x + [1/3]^x + 3 = 9, we can start by simplifying the equation.

Let's rewrite the equation as follows:

(1/2)^x + (1/3)^x + 3 = 9

Now, let's subtract 3 from both sides:

(1/2)^x + (1/3)^x = 6

Next, let's find a common denominator for the fractions:

(3/6)^x + (2/6)^x = 6

Now, let's combine the fractions:

(3^x + 2^x) / 6^x = 6

To simplify further, let's multiply both sides by 6^x:

3^x + 2^x = 6 * 6^x

Now, let's rewrite 6 as 2 * 3:

3^x + 2^x = 2 * 3 * 6^x

Simplifying further:

3^x + 2^x = 2 * 3^(x+1)

Now, let's divide both sides by 3^x:

(3^x + 2^x) / 3^x = (2 * 3^(x+1)) / 3^x

Simplifying:

1 + (2^x / 3^x) = 2 * 3

Now, let's rewrite 2 * 3 as 6:

1 + (2^x / 3^x) = 6

Next, let's subtract 1 from both sides:

2^x / 3^x = 6 - 1

Simplifying:

2^x / 3^x = 5

Now, let's rewrite 5 as 2^2 + 1:

2^x / 3^x = 2^2 + 1

Now, let's rewrite the equation using the same base:

(2^x / 3^x) = (2^2 / 2^0)

Since the bases are the same, the exponents must be equal:

x = 2 - 0

Therefore, the solution to the equation is x = 2.