The value of 'a' when `3^a = 9^b` and `4^((a+b+2)b) = 16^(ab)` is :

Options

A. 2
B. 1
C. 4
D. none of these

Answer

To solve the equations \(3^a = 9^b\) and \(4^{(a+b+2)b} = 16^{(ab)}\), we will follow these steps:

### Step 1: Simplify the first equation

The first equation is \(3^a = 9^b\). We can express \(9\) as \(3^2\):

\[ 9^b = (3^2)^b = 3^{2b} \]

Thus, we can rewrite the equation as:

\[ 3^a = 3^{2b} \]

Since the bases are the same, we can equate the exponents:

\[ a = 2b \quad \text{(1)} \]

### Step 2: Simplify the second equation

The second equation is \(4^{(a+b+2)b} = 16^{(ab)}\). We can express \(16\) as \(4^2\):

\[ 16^{(ab)} = (4^2)^{(ab)} = 4^{2ab} \]

Now, we can rewrite the second equation as:

\[ 4^{(a+b+2)b} = 4^{2ab} \]

Again, since the bases are the same, we can equate the exponents:

\[ (a+b+2)b = 2ab \quad \text{(2)} \]

### Step 3: Substitute equation (1) into equation (2)

From equation (1), we have \(a = 2b\). We can substitute this into equation (2):

\[ (2b + b + 2)b = 2(2b)b \]

This simplifies to:

\[ (3b + 2)b = 4b^2 \]

Expanding the left side gives:

\[ 3b^2 + 2b = 4b^2 \]

### Step 4: Rearrange and solve for \(b\)

Rearranging the equation gives:

\[ 3b^2 + 2b - 4b^2 = 0 \]

This simplifies to:

\[ -b^2 + 2b = 0 \]

Factoring out \(b\):

\[ b(-b + 2) = 0 \]

Thus, we have two solutions:

\[ b = 0 \quad \text{or} \quad b = 2 \]

### Step 5: Find corresponding values of \(a\)

Using \(b = 0\) in equation (1):

\[ a = 2b = 2(0) = 0 \]

Using \(b = 2\) in equation (1):

\[ a = 2b = 2(2) = 4 \]

### Step 6: Conclusion

The values of \(a\) are \(0\) and \(4\). The problem asks for the value of \(a\), so we can conclude that:

\[ \text{The value of } a \text{ can be either } 0 \text{ or } 4. \]