`sqrt((9^((r + (1)/4)) sqrt(3.3^(-r)))/(3.sqrt(3^(-r)))) = k`, then the value of k is

Options

A. 3
B. `3^r`
C. `3^3`
D. `root(r)(3)`

Answer

To solve the equation

\[ \sqrt{\frac{9^{(r + \frac{1}{4})} \sqrt{3} \cdot 3^{-r}}{3 \cdot \sqrt{3^{-r}}}} = k, \]

we will simplify the expression step by step.

### Step 1: Rewrite the expression
We start by rewriting the expression inside the square root.

\[ \sqrt{\frac{9^{(r + \frac{1}{4})} \cdot \sqrt{3} \cdot 3^{-r}}{3 \cdot \sqrt{3^{-r}}}}. \]

### Step 2: Simplify \(9^{(r + \frac{1}{4})}\)
Recall that \(9\) can be expressed as \(3^2\). Thus,

\[ 9^{(r + \frac{1}{4})} = (3^2)^{(r + \frac{1}{4})} = 3^{2(r + \frac{1}{4})} = 3^{2r + \frac{1}{2}}. \]

### Step 3: Substitute back into the expression
Now substitute this back into the expression:

\[ \sqrt{\frac{3^{(2r + \frac{1}{2})} \cdot \sqrt{3} \cdot 3^{-r}}{3 \cdot \sqrt{3^{-r}}}}. \]

### Step 4: Simplify \(\sqrt{3}\) and \(3^{-r}\)
We know that \(\sqrt{3} = 3^{\frac{1}{2}}\) and \(\sqrt{3^{-r}} = 3^{-\frac{r}{2}}\). Thus, we can rewrite the expression as:

\[ \sqrt{\frac{3^{(2r + \frac{1}{2})} \cdot 3^{\frac{1}{2}} \cdot 3^{-r}}{3 \cdot 3^{-\frac{r}{2}}}}. \]

### Step 5: Combine the powers in the numerator
In the numerator, we can combine the powers of \(3\):

\[ 3^{(2r + \frac{1}{2}) + \frac{1}{2} - r} = 3^{(2r - r + \frac{1}{2} + \frac{1}{2})} = 3^{(r + 1)}. \]

### Step 6: Simplify the denominator
In the denominator, we have:

\[ 3 \cdot 3^{-\frac{r}{2}} = 3^{1 - \frac{r}{2}}. \]

### Step 7: Combine the entire expression
Now, we can rewrite the entire expression as:

\[ \sqrt{\frac{3^{(r + 1)}}{3^{(1 - \frac{r}{2})}}} = \sqrt{3^{(r + 1) - (1 - \frac{r}{2})}} = \sqrt{3^{(r + 1 - 1 + \frac{r}{2})}} = \sqrt{3^{(r + \frac{r}{2})}} = \sqrt{3^{(\frac{3r}{2})}}. \]

### Step 8: Simplify the square root
Finally, we simplify the square root:

\[ \sqrt{3^{(\frac{3r}{2})}} = 3^{(\frac{3r}{4})}. \]

### Conclusion
Thus, we find that

\[ k = 3^{(\frac{3r}{4})}. \]