Find the value of `log_9 81 -log_4 32`

Options

A. `1/2`
B. `-3/2`
C. `-1/2`
D. `2`

Answer

To solve the expression \( \log_9 81 - \log_4 32 \), we will break it down step by step.

### Step 1: Simplify \( \log_9 81 \)

We can express \( 81 \) as \( 9^2 \). Therefore, we can rewrite the logarithm:

\[ \log_9 81 = \log_9 (9^2) \]

Using the logarithmic identity \( \log_a (b^n) = n \cdot \log_a b \), we have:

\[ \log_9 (9^2) = 2 \cdot \log_9 9 \]

Since \( \log_9 9 = 1 \):

\[ \log_9 81 = 2 \cdot 1 = 2 \]

### Step 2: Simplify \( \log_4 32 \)

Next, we express \( 32 \) in terms of base \( 4 \). We know that \( 32 = 2^5 \) and \( 4 = 2^2 \). Therefore, we can rewrite the logarithm:

\[ \log_4 32 = \log_{2^2} (2^5) \]

Using the change of base formula \( \log_{a^b} c = \frac{1}{b} \log_a c \):

\[ \log_{2^2} (2^5) = \frac{1}{2} \log_2 (2^5) \]

Again applying the logarithmic identity:

\[ \log_2 (2^5) = 5 \cdot \log_2 2 \]

Since \( \log_2 2 = 1 \):

\[ \log_2 (2^5) = 5 \cdot 1 = 5 \]

Thus, we have:

\[ \log_4 32 = \frac{1}{2} \cdot 5 = \frac{5}{2} \]

### Step 3: Combine the results

Now we can substitute back into the original expression:

\[ \log_9 81 - \log_4 32 = 2 - \frac{5}{2} \]

To perform the subtraction, we convert \( 2 \) into a fraction:

\[ 2 = \frac{4}{2} \]

Now we can subtract:

\[ \frac{4}{2} - \frac{5}{2} = \frac{4 - 5}{2} = \frac{-1}{2} \]

### Final Answer

Thus, the value of \( \log_9 81 - \log_4 32 \) is:

\[ \boxed{-\frac{1}{2}} \]

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