What is the value of `log_(32)27xxlog_(243)8` ?

Options

A. `(log9)/(log4)`
B. `(log3)/(log2)`
C. `log27`
D. None of these

Answer

To solve the expression \( \log_{32} 27 \times \log_{243} 8 \), we will first convert the logarithms into a more manageable form using the change of base formula.

### Step 1: Rewrite the logarithms using the change of base formula
The change of base formula states that:
\[ \log_a b = \frac{\log_c b}{\log_c a} \]
We can use this formula to express both logarithms in terms of natural logarithms or common logarithms (base 10). For simplicity, we will use base 10.

Thus, we have:
\[ \log_{32} 27 = \frac{\log 27}{\log 32} \]
\[ \log_{243} 8 = \frac{\log 8}{\log 243} \]

### Step 2: Substitute the logarithms into the expression
Now, substituting these into our original expression gives:
\[ \log_{32} 27 \times \log_{243} 8 = \frac{\log 27}{\log 32} \times \frac{\log 8}{\log 243} \]

### Step 3: Simplify the logarithms
Next, we can express \(27\), \(32\), \(8\), and \(243\) in terms of powers of prime numbers:
- \(27 = 3^3\)
- \(32 = 2^5\)
- \(8 = 2^3\)
- \(243 = 3^5\)

Using the property of logarithms, \( \log(a^b) = b \log a \), we can rewrite the logarithms:
\[ \log 27 = \log(3^3) = 3 \log 3 \]
\[ \log 32 = \log(2^5) = 5 \log 2 \]
\[ \log 8 = \log(2^3) = 3 \log 2 \]
\[ \log 243 = \log(3^5) = 5 \log 3 \]

### Step 4: Substitute back into the expression
Now substituting these values back into our expression:
\[ \frac{3 \log 3}{5 \log 2} \times \frac{3 \log 2}{5 \log 3} \]

### Step 5: Simplify the expression
Now, we can simplify this expression:
\[ = \frac{3 \log 3 \times 3 \log 2}{5 \log 2 \times 5 \log 3} \]
\[ = \frac{9 \log 3 \log 2}{25 \log 2 \log 3} \]

### Step 6: Cancel out the common terms
The \( \log 3 \) and \( \log 2 \) terms in the numerator and denominator cancel out:
\[ = \frac{9}{25} \]

### Final Answer
Thus, the value of \( \log_{32} 27 \times \log_{243} 8 \) is:
\[ \frac{9}{25} \]