Question

lesson 27 - common and natural logarithms
score: 48/100 answered: 6/11
question 7
use the change-of-base formula to rewrite the following logarithm in terms of both the natural logarithm and the common logarithm. then use a calculator to verify that both expressions are equivalent, and give the calculators approximation rounded to five decimal places.
in terms of the natural logarithm: \\(\log_{5}(63) = \\)
in terms of the common logarithm: \\(\log_{5}(63) = \\)
approximation: \\(\log_{5}(63) \approx \\)
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Explanation:

Step1: Recall change - of - base formula

The change - of - base formula for a logarithm \(\log_{a}(b)\) is \(\log_{a}(b)=\frac{\log_{c}(b)}{\log_{c}(a)}\), where \(c>0,c
eq1\).

Step2: Rewrite in terms of natural logarithm (\(c = e\))

For \(\log_{5}(63)\), using the change - of - base formula with \(c = e\) (natural logarithm, \(\ln\)), we have \(\log_{5}(63)=\frac{\ln(63)}{\ln(5)}\).

Step3: Rewrite in terms of common logarithm (\(c = 10\))

Using the change - of - base formula with \(c = 10\) (common logarithm, \(\log\)), we get \(\log_{5}(63)=\frac{\log(63)}{\log(5)}\).

Step4: Calculate the approximation

First, calculate \(\frac{\ln(63)}{\ln(5)}\):
\(\ln(63)\approx4.143134727\), \(\ln(5)\approx1.609437912\)
\(\frac{\ln(63)}{\ln(5)}\approx\frac{4.143134727}{1.609437912}\approx2.57496\) (rounded to five decimal places)

Or calculate \(\frac{\log(63)}{\log(5)}\):
\(\log(63)\approx1.79934\), \(\log(5)\approx0.69897\)
\(\frac{\log(63)}{\log(5)}\approx\frac{1.79934}{0.69897}\approx2.57496\) (rounded to five decimal places)

Answer:

In terms of the natural logarithm: \(\frac{\ln(63)}{\ln(5)}\)
In terms of the common logarithm: \(\frac{\log(63)}{\log(5)}\)
Approximation: \(2.57496\)