To solve this equation, we need to expand the left side of the equation using the binomial theorem.

The binomial theorem states that for any real numbers a and b, and any positive integer n, we have:

(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n

where C(n, k) is the binomial coefficient, given by:

C(n, k) = n! / (k! * (n-k)!)

In our case, a = 2x and b = 1, and n = 5. So, we can expand (2x + 1)^5 as follows:

(2x + 1)^5 = C(5, 0) * (2x)^5 * 1^0 + C(5, 1) * (2x)^4 * 1^1 + C(5, 2) * (2x)^3 * 1^2 + C(5, 3) * (2x)^2 * 1^3 + C(5, 4) * (2x)^1 * 1^4 + C(5, 5) * (2x)^0 * 1^5

Simplifying each term:

= 1 * (2x)^5 * 1 + 5 * (2x)^4 * 1 + 10 * (2x)^3 * 1^2 + 10 * (2x)^2 * 1^3 + 5 * (2x)^1 * 1^4 + 1 * (2x)^0 * 1^5

= (2x)^5 + 5 * (2x)^4 + 10 * (2x)^3 + 10 * (2x)^2 + 5 * (2x) + 1

Now, we can substitute this expression into the equation:

(2x)^5 + 5 * (2x)^4 + 10 * (2x)^3 + 10 * (2x)^2 + 5 * (2x) + 1 = 3^3 * 9

Simplifying the right side:

(2x)^5 + 5 * (2x)^4 + 10 * (2x)^3 + 10 * (2x)^2 + 5 * (2x) + 1 = 27 * 9

(2x)^5 + 5 * (2x)^4 + 10 * (2x)^3 + 10 * (2x)^2 + 5 * (2x) + 1 = 243

Now, we can solve this equation for x. However, it is a quintic equation, which does not have a general algebraic solution. We would need to use numerical methods or approximation techniques to find the value of x that satisfies this equation.