Questions: Evaluate each expression. (a) log base 7 of 1 = (b) log base 9 of 1/81 =

Transcript text: Evaluate each expression. (a) $\log _{7} 1=$ $\square$ (b) $\log _{9} \frac{1}{81}=$ $\square$

Solution

Solution Steps

Solution Approach

To evaluate logarithmic expressions, we use the property that $\log_b a = c$ means $b^c = a$. For each expression: (a) $\log_7 1$: Any number raised to the power of 0 is 1, so $\log_7 1 = 0$. (b) $\log_9 \frac{1}{81}$: Recognize that $\frac{1}{81} = 9^{-2}$, so $\log_9 \frac{1}{81} = -2$.

Step 1: Evaluate $ \log_7 1 $

Using the property of logarithms, we know that $ \log_b a = c $ implies $ b^c = a $. For $ \log_7 1 $, we can express this as: \[ 7^c = 1 \] The only exponent $ c $ that satisfies this equation is $ c = 0 $. Therefore, we have: \[ \log_7 1 = 0 \]

Step 2: Evaluate $ \log_9 \frac{1}{81} $

We can rewrite $ \frac{1}{81} $ as $ 9^{-2} $. Thus, we can express the logarithm as: \[ \log_9 \frac{1}{81} = \log_9 (9^{-2}) \] Using the property of logarithms that states $ \log_b (b^c) = c $, we find: \[ \log_9 (9^{-2}) = -2 \]

Final Answer

The answers to the logarithmic expressions are: \[ \log_7 1 = 0 \quad \text{and} \quad \log_9 \frac{1}{81} = -2 \] Thus, the final boxed answers are: \[ \boxed{0} \quad \text{and} \quad \boxed{-2} \]

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