Prove that `log_(3) 81- 4 =0`

Options

A. `0ltxlt10`
B. `0ltxlt10^10`
C. `0ltxlt10^(1//10)`
D. None of these

Answer

To prove that \( \log_{3} 81 - 4 = 0 \), we will follow these steps:

### Step 1: Rewrite the logarithm
We start with the left-hand side (LHS):
\[ \text{LHS} = \log_{3} 81 - 4 \]

### Step 2: Express 81 as a power of 3
We know that \( 81 \) can be expressed as \( 3^4 \) because:
\[ 81 = 3 \times 3 \times 3 \times 3 = 3^4 \]

### Step 3: Substitute into the logarithm
Now we can substitute \( 81 \) with \( 3^4 \) in the logarithm:
\[ \text{LHS} = \log_{3} (3^4) - 4 \]

### Step 4: Apply the power rule of logarithms
Using the property of logarithms that states \( \log_{a} (b^n) = n \cdot \log_{a} b \), we can simplify:
\[ \text{LHS} = 4 \cdot \log_{3} 3 - 4 \]

### Step 5: Simplify using the identity of logarithms
We know that \( \log_{3} 3 = 1 \) because the logarithm of a number to its own base is always 1:
\[ \text{LHS} = 4 \cdot 1 - 4 \]

### Step 6: Final simplification
Now we can simplify further:
\[ \text{LHS} = 4 - 4 = 0 \]

### Conclusion
Thus, we have shown that:
\[ \log_{3} 81 - 4 = 0 \]
This proves the statement.