If log 2 =x log 3=y and log 7 =z then the value of `log(4xx root 3(63))`

Options

A. (-2x+2/3y+1/3z)
B. 2x+2/3y+1/3z
C. 2x+2/3y-1/3z
D. 2x-2/3 y+1/3 z

Answer

To solve the problem, we need to find the value of \( \log(4 \cdot \sqrt[3]{63}) \) given that \( \log 2 = x \), \( \log 3 = y \), and \( \log 7 = z \).

### Step 1: Rewrite the expression using logarithmic properties
We can rewrite \( \log(4 \cdot \sqrt[3]{63}) \) using the property \( \log(m \cdot n) = \log m + \log n \):
\[ \log(4 \cdot \sqrt[3]{63}) = \log 4 + \log(\sqrt[3]{63}) \]

### Step 2: Simplify \( \log 4 \)
Since \( 4 = 2^2 \), we can use the property \( \log(m^n) = n \cdot \log m \):
\[ \log 4 = \log(2^2) = 2 \log 2 = 2x \]

### Step 3: Simplify \( \log(\sqrt[3]{63}) \)
The cube root can be expressed as a power:
\[ \sqrt[3]{63} = 63^{1/3} \]
Thus, we can write:
\[ \log(\sqrt[3]{63}) = \log(63^{1/3}) = \frac{1}{3} \log 63 \]

### Step 4: Simplify \( \log 63 \)
Next, we express \( 63 \) in terms of its prime factors:
\[ 63 = 7 \cdot 9 = 7 \cdot 3^2 \]
Using the property of logarithms:
\[ \log 63 = \log(7 \cdot 3^2) = \log 7 + \log(3^2) = \log 7 + 2 \log 3 = z + 2y \]

### Step 5: Substitute back into the expression
Now we can substitute \( \log 63 \) back into our expression for \( \log(\sqrt[3]{63}) \):
\[ \log(\sqrt[3]{63}) = \frac{1}{3} \log 63 = \frac{1}{3}(z + 2y) = \frac{1}{3}z + \frac{2}{3}y \]

### Step 6: Combine the results
Now we can combine \( \log 4 \) and \( \log(\sqrt[3]{63}) \):
\[ \log(4 \cdot \sqrt[3]{63}) = 2x + \left(\frac{1}{3}z + \frac{2}{3}y\right) \]
This simplifies to:
\[ \log(4 \cdot \sqrt[3]{63}) = 2x + \frac{2}{3}y + \frac{1}{3}z \]

### Final Answer
Thus, the value of \( \log(4 \cdot \sqrt[3]{63}) \) is:
\[ \boxed{2x + \frac{2}{3}y + \frac{1}{3}z} \]