Express in terms of log 2 and log 3 :
` log ""(sqrt(54) xx ""^(3) sqrt(243))`

Answer

To express \( \log \left( \sqrt{54} \times \left( \sqrt{243} \right)^{\frac{1}{3}} \right) \) in terms of \( \log 2 \) and \( \log 3 \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \log \left( \sqrt{54} \times \left( \sqrt{243} \right)^{\frac{1}{3}} \right) \]
Using the property of logarithms that states \( \log(a \times b) = \log a + \log b \), we can separate the logarithm:
\[ \log \left( \sqrt{54} \right) + \log \left( \left( \sqrt{243} \right)^{\frac{1}{3}} \right) \]

### Step 2: Simplify each term
Now we simplify each term:
1. For \( \log \left( \sqrt{54} \right) \):
\[ \log \left( \sqrt{54} \right) = \log \left( 54^{\frac{1}{2}} \right) = \frac{1}{2} \log 54 \]
2. For \( \log \left( \left( \sqrt{243} \right)^{\frac{1}{3}} \right) \):
\[ \log \left( \left( \sqrt{243} \right)^{\frac{1}{3}} \right) = \frac{1}{3} \log \left( \sqrt{243} \right) = \frac{1}{3} \cdot \frac{1}{2} \log 243 = \frac{1}{6} \log 243 \]

Combining these, we have:
\[ \frac{1}{2} \log 54 + \frac{1}{6} \log 243 \]

### Step 3: Express 54 and 243 in terms of their prime factors
Next, we express \( 54 \) and \( 243 \) in terms of their prime factors:
- \( 54 = 2 \times 3^3 \)
- \( 243 = 3^5 \)

Thus, we can rewrite the logarithms:
1. For \( \log 54 \):
\[ \log 54 = \log (2 \times 3^3) = \log 2 + \log (3^3) = \log 2 + 3 \log 3 \]
2. For \( \log 243 \):
\[ \log 243 = \log (3^5) = 5 \log 3 \]

### Step 4: Substitute back into the expression
Now substituting these back into our expression:
\[ \frac{1}{2} (\log 2 + 3 \log 3) + \frac{1}{6} (5 \log 3) \]

### Step 5: Simplify the expression
Distributing the coefficients:
\[ \frac{1}{2} \log 2 + \frac{3}{2} \log 3 + \frac{5}{6} \log 3 \]

Now, combine the terms involving \( \log 3 \):
\[ \frac{1}{2} \log 2 + \left( \frac{3}{2} + \frac{5}{6} \right) \log 3 \]

To combine \( \frac{3}{2} \) and \( \frac{5}{6} \), we find a common denominator (which is 6):
\[ \frac{3}{2} = \frac{9}{6} \]
So,
\[ \frac{9}{6} + \frac{5}{6} = \frac{14}{6} = \frac{7}{3} \]

### Final Expression
Thus, the final expression is:
\[ \frac{1}{2} \log 2 + \frac{7}{3} \log 3 \]