Find the value of `(log)_(2sqrt(3))1728.`

Options

A. 6
B. 8
C. 3
D. 5

Answer

To find the value of \( \log_{2\sqrt{3}} 1728 \), we can follow these steps:

### Step 1: Use the Change of Base Formula
We can use the change of base formula for logarithms, which states:
\[ \log_a b = \frac{\log_e b}{\log_e a} \]
Thus, we can rewrite our expression as:
\[ \log_{2\sqrt{3}} 1728 = \frac{\log 1728}{\log (2\sqrt{3})} \]

### Step 2: Factor 1728
Next, we need to factor 1728. We can express 1728 as:
\[ 1728 = 2^6 \times 3^3 \]

### Step 3: Apply Logarithmic Properties
Using the properties of logarithms, we can write:
\[ \log 1728 = \log (2^6 \times 3^3) = \log (2^6) + \log (3^3) \]
Using the power rule of logarithms, this becomes:
\[ \log 1728 = 6 \log 2 + 3 \log 3 \]

### Step 4: Simplify the Denominator
Now, we simplify the denominator \( \log (2\sqrt{3}) \):
\[ \log (2\sqrt{3}) = \log (2) + \log (\sqrt{3}) = \log (2) + \frac{1}{2} \log (3) \]

### Step 5: Substitute Back into the Expression
Now, substituting back into our expression, we have:
\[ \log_{2\sqrt{3}} 1728 = \frac{6 \log 2 + 3 \log 3}{\log 2 + \frac{1}{2} \log 3} \]

### Step 6: Multiply by 2 to Clear the Denominator
To simplify further, we can multiply the numerator and denominator by 2:
\[ \log_{2\sqrt{3}} 1728 = \frac{2(6 \log 2 + 3 \log 3)}{2 \log 2 + \log 3} \]
This gives us:
\[ = \frac{12 \log 2 + 6 \log 3}{2 \log 2 + \log 3} \]

### Step 7: Factor Out Common Terms
Now we can factor out \( 2 \) from the numerator:
\[ = \frac{6(2 \log 2 + \log 3)}{2 \log 2 + \log 3} \]

### Step 8: Simplify
Since \( 2 \log 2 + \log 3 \) cancels out, we are left with:
\[ \log_{2\sqrt{3}} 1728 = 6 \]

### Final Answer
Thus, the value of \( \log_{2\sqrt{3}} 1728 \) is:
\[ \boxed{6} \]