Sure! Let's work through each of these logarithm problems step-by-step, using the properties of logarithms to simplify the expressions.
### 1. [tex]\(\log_3 9 + \log_3 27\)[/tex]
Using the property of logarithms that states [tex]\(\log_b(m) + \log_b(n) = \log_b(m \cdot n)\)[/tex]:
[tex]\[
\log_3 9 + \log_3 27 = \log_3 (9 \times 27)
\][/tex]
Next, we multiply 9 and 27:
[tex]\[
9 \times 27 = 243
\][/tex]
So,
[tex]\[
\log_3 9 + \log_3 27 = \log_3 243 = \boxed{5}
\][/tex]
### 2. [tex]\(\log_2 8 - \log_2 4\)[/tex]
Using the property of logarithms that states [tex]\(\log_b(m) - \log_b(n) = \log_b \left(\frac{m}{n}\right)\)[/tex]:
[tex]\[
\log_2 8 - \log_2 4 = \log_2 \left(\frac{8}{4}\right)
\][/tex]
Next, we divide 8 by 4:
[tex]\[
\frac{8}{4} = 2
\][/tex]
So,
[tex]\[
\log_2 8 - \log_2 4 = \log_2 2 = \boxed{1}
\][/tex]
### 3. [tex]\(\log_5 5 (5^{\frac{1}{3}})\)[/tex]
Using the property of logarithms that states [tex]\(\log_b(m \cdot n) = \log_b m + \log_b n\)[/tex]:
[tex]\[
\log_5 5 + \log_5 (5^{\frac{1}{3}})
\][/tex]
First, recall that [tex]\(\log_b b = 1\)[/tex], so:
[tex]\[
\log_5 5 = 1
\][/tex]
Next, use the property of logarithms that states [tex]\(\log_b (a^c) = c \cdot \log_b a\)[/tex]:
[tex]\[
\log_5 (5^{\frac{1}{3}}) = \frac{1}{3} \log_5 5
\][/tex]
Since [tex]\(\log_5 5 = 1\)[/tex]:
[tex]\[
\log_5 (5^{\frac{1}{3}}) = \frac{1}{3} \cdot 1 = \frac{1}{3}
\][/tex]
Putting it all together:
[tex]\[
\log_5 5 + \log_5 (5^{\frac{1}{3}}) = 1 + \frac{1}{3} = \boxed{1.3333}
\][/tex]
So, the final evaluations are:
1. [tex]\(\log_3 9 + \log_3 27 = 5\)[/tex]
2. [tex]\(\log_2 8 - \log_2 4 = 1\)[/tex]
3. [tex]\(\log_5 5 (5^{\frac{1}{3}}) = 1.3333\)[/tex]
