2. Evaluate. (1) $$ \log _{s} 1 $$ (2) $$ \log _{4} 4 $$ (3) $$ \log _{7} 7^{3} $$ (4) $$ b^{\log _{8} 3} $$ (3) $$ \log _{25} 5^{3} $$ (4) $$ 16^{\log _{4} 8} $$

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The logarithm of any number to the base $$ s $$ where $$ s > 0 $$ is equal to 0. Therefore, $$ \log_{s} 1 = 0 $$.
The logarithm of a number to its own base is always 1. Thus, $$ \log_{4} 4 = 1 $$.
Using the property of logarithms, $$ \log_{b} b^{k} = k $$. Therefore, $$ \log_{7} 7^{3} = 3 $$.
This expression cannot be simplified further without specific values for $$ b $$ and $$ 3 $$. It remains as $$ b^{\log_{8} 3} $$.
Using the change of base property, we can express $$ 5^{3} $$ as $$ (5^{2})^{1.5} $$ and apply the logarithm property: $$ \log_{b} a^{k} = k \cdot \log_{b} a $$. Thus, $$ \log_{25} 5^{3} = 3 \cdot \log_{25} 5 $$. Since $$ 25 = 5^{2} $$, we have $$ \log_{25} 5 = \frac{1}{2} $$. Therefore, $$ \log_{25} 5^{3} = 3 \cdot \frac{1}{2} = \frac{3}{2} $$.
Using the property $$ a^{\log_{b} c} = c^{\log_{b} a} $$, we can rewrite $$ 16^{\log_{4} 8} $$ as $$ 8^{\log_{4} 16} $$. Since $$ 16 = 4^{2} $$, we have $$ \log_{4} 16 = 2 $$. Thus, $$ 8^{\log_{4} 16} = 8^{2} = 64 $$.
1. $$ \log_{s} 1 = 0 $$ 2. $$ \log_{4} 4 = 1 $$ 3. $$ \log_{7} 7^{3} = 3 $$ 4. $$ b^{\log_{8} 3} $$ (remains as is) 5. $$ \log_{25} 5^{3} = \frac{3}{2} $$ 6. $$ 16^{\log_{4} 8} = 64 $$

Evaluate.
(1)
(2)
(3)
(4)
(3)
(4)

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