Sure, I'd be happy to help you with these expressions.
(a) [tex]\(\log_2\left(\frac{1}{8}\right) = \square\)[/tex]
We rewrite [tex]\(\frac{1}{8}\)[/tex] as [tex]\(8^{-1}\)[/tex]. Since [tex]\(8 = 2^3\)[/tex], it follows that:
[tex]\[
\frac{1}{8} = 2^{-3}
\][/tex]
Therefore, we have:
[tex]\[
\log_2\left(\frac{1}{8}\right) = \log_2(2^{-3})
\][/tex]
Using the property of logarithms, [tex]\(\log_b(a^c) = c \log_b(a)\)[/tex], we get:
[tex]\[
\log_2(2^{-3}) = -3 \log_2(2)
\][/tex]
Since [tex]\(\log_2(2) = 1\)[/tex], we find:
[tex]\[
-3 \log_2(2) = -3 \cdot 1 = -3
\][/tex]
Thus,
[tex]\[
\log_2\left(\frac{1}{8}\right) = -3
\][/tex]
(b) [tex]\(\log_{19} 1 = \square\)[/tex]
By the definition of logarithms, [tex]\(\log_b(1) = 0\)[/tex] for any base [tex]\(b > 0\)[/tex] because any number raised to the power of 0 is 1. Hence:
[tex]\[
\log_{19} 1 = 0
\][/tex]
(c) [tex]\(\log_3 \sqrt{27} = \square\)[/tex]
First, express [tex]\(\sqrt{27}\)[/tex] in terms of powers of 3. We know [tex]\(27 = 3^3\)[/tex], so:
[tex]\[
\sqrt{27} = \sqrt{3^3} = (3^3)^{1/2}
\][/tex]
Using the property of exponents, [tex]\((a^m)^n = a^{m \cdot n}\)[/tex], we get:
[tex]\[
(3^3)^{1/2} = 3^{3/2}
\][/tex]
Therefore,
[tex]\[
\log_3 \sqrt{27} = \log_3(3^{3/2})
\][/tex]
Using the property of logarithms, [tex]\(\log_b(a^c) = c \log_b(a)\)[/tex], we have:
[tex]\[
\log_3(3^{3/2}) = \frac{3}{2} \log_3(3)
\][/tex]
Since [tex]\(\log_3(3) = 1\)[/tex], we get:
[tex]\[
\frac{3}{2} \log_3(3) = \frac{3}{2} \cdot 1 = \frac{3}{2} = 1.5
\][/tex]
Thus,
[tex]\[
\log_3 \sqrt{27} = 1.5
\][/tex]
(d) [tex]\(3^{\log_3 8} = \square\)[/tex]
Using the property of logarithms and exponents where [tex]\(a^{\log_a(b)} = b\)[/tex], we have:
[tex]\[
3^{\log_3 8} = 8
\][/tex]
Putting it all together, the solutions are:
(a) [tex]\(\log_2\left(\frac{1}{8}\right) = -3\)[/tex]
(b) [tex]\(\log_{19} 1 = 0\)[/tex]
(c) [tex]\(\log_3 \sqrt{27} = 1.5\)[/tex]
(d) [tex]\(3^{\log_3 8} = 8\)[/tex]
