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Answer :
We want to find the approximate value of
[tex]$$
\log_4 128.
$$[/tex]
Step 1: We use the change of base formula, which states that for any positive numbers [tex]\(a\)[/tex], [tex]\(b\)[/tex], and base [tex]\(c\)[/tex] (with [tex]\(c \neq 1\)[/tex]),
[tex]$$
\log_a b = \frac{\log b}{\log a}.
$$[/tex]
Step 2: Applying this to our problem, we have
[tex]$$
\log_4 128 = \frac{\log 128}{\log 4}.
$$[/tex]
Step 3: Substitute the given values:
[tex]$$
\log 128 \approx 2.1 \quad \text{and} \quad \log 4 \approx 0.6.
$$[/tex]
Step 4: Compute the quotient:
[tex]$$
\log_4 128 \approx \frac{2.1}{0.6} \approx 3.5.
$$[/tex]
Thus, the approximate value of [tex]\( \log_4 128 \)[/tex] is [tex]\( \boxed{3.5} \)[/tex].
[tex]$$
\log_4 128.
$$[/tex]
Step 1: We use the change of base formula, which states that for any positive numbers [tex]\(a\)[/tex], [tex]\(b\)[/tex], and base [tex]\(c\)[/tex] (with [tex]\(c \neq 1\)[/tex]),
[tex]$$
\log_a b = \frac{\log b}{\log a}.
$$[/tex]
Step 2: Applying this to our problem, we have
[tex]$$
\log_4 128 = \frac{\log 128}{\log 4}.
$$[/tex]
Step 3: Substitute the given values:
[tex]$$
\log 128 \approx 2.1 \quad \text{and} \quad \log 4 \approx 0.6.
$$[/tex]
Step 4: Compute the quotient:
[tex]$$
\log_4 128 \approx \frac{2.1}{0.6} \approx 3.5.
$$[/tex]
Thus, the approximate value of [tex]\( \log_4 128 \)[/tex] is [tex]\( \boxed{3.5} \)[/tex].
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