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Answer :
Sure, let's solve the problem step-by-step.
To find the value of [tex]\(\log_4 128\)[/tex], we can use the change of base formula. The change of base formula is:
[tex]\[
\log_a b = \frac{\log_c b}{\log_c a}
\][/tex]
In simpler terms, to find [tex]\(\log_4 128\)[/tex], we can rewrite it using common logarithms (log base 10):
[tex]\[
\log_4 128 = \frac{\log 128}{\log 4}
\][/tex]
We are given the approximate values:
[tex]\[
\log 128 \approx 2.1
\][/tex]
[tex]\[
\log 4 \approx 0.6
\][/tex]
Now, substitute these values into the formula:
[tex]\[
\log_4 128 = \frac{2.1}{0.6}
\][/tex]
Perform the division:
[tex]\[
\log_4 128 \approx \frac{2.1}{0.6} \approx 3.5
\][/tex]
Therefore, the approximate value of [tex]\(\log_4 128\)[/tex] is [tex]\( \boxed{3.5} \)[/tex].
To find the value of [tex]\(\log_4 128\)[/tex], we can use the change of base formula. The change of base formula is:
[tex]\[
\log_a b = \frac{\log_c b}{\log_c a}
\][/tex]
In simpler terms, to find [tex]\(\log_4 128\)[/tex], we can rewrite it using common logarithms (log base 10):
[tex]\[
\log_4 128 = \frac{\log 128}{\log 4}
\][/tex]
We are given the approximate values:
[tex]\[
\log 128 \approx 2.1
\][/tex]
[tex]\[
\log 4 \approx 0.6
\][/tex]
Now, substitute these values into the formula:
[tex]\[
\log_4 128 = \frac{2.1}{0.6}
\][/tex]
Perform the division:
[tex]\[
\log_4 128 \approx \frac{2.1}{0.6} \approx 3.5
\][/tex]
Therefore, the approximate value of [tex]\(\log_4 128\)[/tex] is [tex]\( \boxed{3.5} \)[/tex].
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