If `root3((3^(6) xx 4^(3) xx 2^(6))/(8^(9) xx2^(3)))=(3/8)^(k)` ,then k = _______ .

Answer

To solve the equation

\[ \sqrt[3]{\frac{3^{6} \times 4^{3} \times 2^{6}}{8^{9} \times 2^{3}}} = \left(\frac{3}{8}\right)^{k}, \]

we will follow these steps:

### Step 1: Rewrite the terms
First, we rewrite \(4^3\) and \(8^9\) in terms of base \(2\):
- \(4^3 = (2^2)^3 = 2^{6}\)
- \(8^9 = (2^3)^9 = 2^{27}\)

Now, substituting these into the equation gives us:

\[ \sqrt[3]{\frac{3^{6} \times 2^{6} \times 2^{6}}{2^{27} \times 2^{3}}} \]

### Step 2: Simplify the fraction
Next, we simplify the fraction inside the cube root:

The numerator becomes:
\[ 3^{6} \times 2^{6} \times 2^{6} = 3^{6} \times 2^{12} \]

The denominator becomes:
\[ 2^{27} \times 2^{3} = 2^{30} \]

Thus, we have:

\[ \sqrt[3]{\frac{3^{6} \times 2^{12}}{2^{30}}} \]

### Step 3: Further simplify the fraction
Now, we can simplify the fraction:

\[ \frac{2^{12}}{2^{30}} = 2^{12 - 30} = 2^{-18} \]

So the expression becomes:

\[ \sqrt[3]{3^{6} \times 2^{-18}} = \sqrt[3]{3^{6}} \times \sqrt[3]{2^{-18}} \]

### Step 4: Apply the cube root
Now we apply the cube root:

\[ \sqrt[3]{3^{6}} = 3^{6/3} = 3^{2} \]
\[ \sqrt[3]{2^{-18}} = 2^{-18/3} = 2^{-6} \]

Thus, we have:

\[ 3^{2} \times 2^{-6} = \frac{3^{2}}{2^{6}} \]

### Step 5: Rewrite in terms of \(\frac{3}{8}\)
Now we can express \(2^{6}\) as \(8^{2}\):

\[ \frac{3^{2}}{2^{6}} = \frac{3^{2}}{(2^{3})^{2}} = \frac{3^{2}}{8^{2}} \]

### Step 6: Set equal to the original expression
Now we can set this equal to the right side of the original equation:

\[ \frac{3^{2}}{8^{2}} = \left(\frac{3}{8}\right)^{k} \]

### Step 7: Compare the powers
Since the bases are the same, we can equate the exponents:

\[ 2 = k \]

Thus, the value of \(k\) is:

\[ \boxed{2} \]

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