Evaluate the following expressions without using a calculator.

(a) [tex]\log _2\left(\frac{1}{8}\right) = \square[/tex]
(b) [tex]\log _7 1 = \square[/tex]
(c) [tex]\log _4 \sqrt{1024} = \square[/tex]
(d) [tex]9^{\log _9 9} = \square[/tex]



Answer :

Sure, let's evaluate each expression step-by-step:

### (a) [tex]\(\log_2\left(\frac{1}{8}\right)\)[/tex]

To solve [tex]\(\log_2\left(\frac{1}{8}\right)\)[/tex], we can use the property of logarithms that states [tex]\(\log_b\left(\frac{1}{a}\right) = -\log_b(a)\)[/tex]. So, we have:
[tex]\[ \log_2\left(\frac{1}{8}\right) = -\log_2(8) \][/tex]

Next, we need to evaluate [tex]\(\log_2(8)\)[/tex]. We know that [tex]\(8\)[/tex] can be expressed as a power of [tex]\(2\)[/tex]:
[tex]\[ 8 = 2^3 \][/tex]
Therefore,
[tex]\[ \log_2(8) = \log_2(2^3) = 3 \][/tex]
So,
[tex]\[ \log_2\left(\frac{1}{8}\right) = -3 \][/tex]

Thus, [tex]\(\log_2\left(\frac{1}{8}\right) = -3\)[/tex].

### (b) [tex]\(\log_7 1\)[/tex]

By the properties of logarithms, we know that for any base [tex]\(b\)[/tex], [tex]\(\log_b(1)\)[/tex] is always [tex]\(0\)[/tex] because any number raised to the power of [tex]\(0\)[/tex] is [tex]\(1\)[/tex]:
[tex]\[ \log_7 1 = 0 \][/tex]

So, [tex]\(\log_7 1 = 0\)[/tex].

### (c) [tex]\(\log_4 \sqrt{1024}\)[/tex]

To solve [tex]\(\log_4 \sqrt{1024}\)[/tex], we first express [tex]\(\sqrt{1024}\)[/tex] in exponential form:
[tex]\[ \sqrt{1024} = 1024^{\frac{1}{2}} \][/tex]
Thus,
[tex]\[ \log_4 \sqrt{1024} = \log_4\left(1024^{\frac{1}{2}}\right) \][/tex]

Using the logarithm property [tex]\(\log_b(a^c) = c \cdot \log_b(a)\)[/tex], we get:
[tex]\[ \log_4\left(1024^{\frac{1}{2}}\right) = \frac{1}{2} \cdot \log_4(1024) \][/tex]

Next, we need to evaluate [tex]\(\log_4(1024)\)[/tex]. We know that [tex]\(1024\)[/tex] can be expressed as a power of [tex]\(4\)[/tex]:
[tex]\[ 1024 = 4^5 \][/tex]
Therefore,
[tex]\[ \log_4(1024) = \log_4(4^5) = 5 \][/tex]
So,
[tex]\[ \log_4 \sqrt{1024} = \frac{1}{2} \cdot 5 = 2.5 \][/tex]

Thus, [tex]\(\log_4 \sqrt{1024} = 2.5\)[/tex].

### (d) [tex]\(9^{\log_9 9}\)[/tex]

To solve [tex]\(9^{\log_9 9}\)[/tex], we use the property of exponents and logarithms that states [tex]\(b^{\log_b(x)} = x\)[/tex]. In this case:
[tex]\[ 9^{\log_9 9} = 9 \][/tex]

Thus, [tex]\(9^{\log_9 9} = 9\)[/tex].

In summary, the evaluated expressions are:
[tex]\[ \boxed{-3}, \boxed{0}, \boxed{2.5}, \boxed{9} \][/tex]