If the coefficients of `x^2 and x^3` in the expansion of `(3 + ax)^(9)` are the same, find the value of a.

Answer

To find the value of \( a \) such that the coefficients of \( x^2 \) and \( x^3 \) in the expansion of \( (3 + ax)^9 \) are equal, we can follow these steps:

### Step 1: Identify the General Term
The general term in the binomial expansion of \( (3 + ax)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} (3)^{n-r} (ax)^r \]
For our case, \( n = 9 \), so the general term becomes:
\[ T_{r+1} = \binom{9}{r} (3)^{9-r} (ax)^r = \binom{9}{r} (3)^{9-r} a^r x^r \]

### Step 2: Find the Coefficient of \( x^2 \)
To find the coefficient of \( x^2 \), we set \( r = 2 \):
\[ \text{Coefficient of } x^2 = \binom{9}{2} (3)^{9-2} a^2 \]
Calculating this:
\[ \text{Coefficient of } x^2 = \binom{9}{2} (3)^7 a^2 = \frac{9 \times 8}{2 \times 1} \cdot 2187 \cdot a^2 = 36 \cdot 2187 \cdot a^2 \]

### Step 3: Find the Coefficient of \( x^3 \)
To find the coefficient of \( x^3 \), we set \( r = 3 \):
\[ \text{Coefficient of } x^3 = \binom{9}{3} (3)^{9-3} a^3 \]
Calculating this:
\[ \text{Coefficient of } x^3 = \binom{9}{3} (3)^6 a^3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \cdot 729 \cdot a^3 = 84 \cdot 729 \cdot a^3 \]

### Step 4: Set the Coefficients Equal
Since the coefficients of \( x^2 \) and \( x^3 \) are equal, we can set them equal to each other:
\[ 36 \cdot 2187 \cdot a^2 = 84 \cdot 729 \cdot a^3 \]

### Step 5: Simplify the Equation
We can simplify this equation:
\[ 36 \cdot 2187 = 84 \cdot 729 \cdot a \]
Calculating \( 36 \cdot 2187 \) and \( 84 \cdot 729 \):
\[ 36 \cdot 2187 = 78732 \]
\[ 84 \cdot 729 = 61236 \]
So we have:
\[ 78732 = 61236 \cdot a \]

### Step 6: Solve for \( a \)
Now, we can solve for \( a \):
\[ a = \frac{78732}{61236} \]
Calculating this gives:
\[ a = \frac{78732 \div 61236}{61236 \div 61236} = \frac{13}{10} = 1.3 \]

### Final Answer
Thus, the value of \( a \) is:
\[ \boxed{\frac{13}{10}} \]