Find the value of `log_(32) 2^8 +log_(243) 3^7 -log_(36) 1296`

Options

A. 3
B. 2
C. 1
D. 0

Answer

To solve the expression \( \log_{32} 2^8 + \log_{243} 3^7 - \log_{36} 1296 \), we will use the properties of logarithms. Let's break it down step by step.

### Step 1: Simplify \( \log_{32} 2^8 \)
We know that \( 32 = 2^5 \). Therefore, we can rewrite the logarithm as follows:
\[ \log_{32} 2^8 = \log_{2^5} 2^8 \]
Using the property of logarithms \( \log_{a^m} b^n = \frac{n}{m} \log_a b \):
\[ \log_{2^5} 2^8 = \frac{8}{5} \log_2 2 \]
Since \( \log_2 2 = 1 \):
\[ \log_{32} 2^8 = \frac{8}{5} \cdot 1 = \frac{8}{5} \]

### Step 2: Simplify \( \log_{243} 3^7 \)
Next, we know that \( 243 = 3^5 \). Thus, we can rewrite this logarithm as:
\[ \log_{243} 3^7 = \log_{3^5} 3^7 \]
Using the same property:
\[ \log_{3^5} 3^7 = \frac{7}{5} \log_3 3 \]
Since \( \log_3 3 = 1 \):
\[ \log_{243} 3^7 = \frac{7}{5} \cdot 1 = \frac{7}{5} \]

### Step 3: Simplify \( \log_{36} 1296 \)
Now, we know that \( 1296 = 6^4 \) and \( 36 = 6^2 \). Therefore, we can rewrite this logarithm as:
\[ \log_{36} 1296 = \log_{6^2} 6^4 \]
Using the logarithmic property:
\[ \log_{6^2} 6^4 = \frac{4}{2} \log_6 6 \]
Since \( \log_6 6 = 1 \):
\[ \log_{36} 1296 = \frac{4}{2} \cdot 1 = 2 \]

### Step 4: Combine the results
Now we can combine all the results:
\[ \log_{32} 2^8 + \log_{243} 3^7 - \log_{36} 1296 = \frac{8}{5} + \frac{7}{5} - 2 \]
First, add the two fractions:
\[ \frac{8}{5} + \frac{7}{5} = \frac{15}{5} = 3 \]
Now subtract \( 2 \):
\[ 3 - 2 = 1 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{1} \]