Algebra 2 Worksheets | Exponential and Logarithmic Functions ... - Free Printable

Problem Overview:

The task involves using the Change of Base Formula to approximate logarithms to the nearest hundredth. The Change of Base Formula is given by:

\[
\log_b a = \frac{\log_k a}{\log_k b}
\]

where \( k \) can be any positive base (commonly 10 or \( e \)). For this problem, we will use the natural logarithm (\( \ln \)) as the base \( k \).

Solution Approach:

1. Identify the base and argument for each logarithm.
2. Apply the Change of Base Formula:
\[
\log_b a = \frac{\ln a}{\ln b}
\]
3. Use a calculator to compute the natural logarithms (\( \ln \)).
4. Perform the division and round the result to the nearest hundredth.

Detailed Solutions:

#### 1) \( \ln 139 \)
This is a natural logarithm, so no change of base is needed.
\[
\ln 139 \approx 4.934
\]

#### 2) \( \log_8 44.3 \)
Using the Change of Base Formula:
\[
\log_8 44.3 = \frac{\ln 44.3}{\ln 8}
\]
Calculate:
\[
\ln 44.3 \approx 3.791, \quad \ln 8 \approx 2.079
\]
\[
\log_8 44.3 = \frac{3.791}{2.079} \approx 1.823
\]
Rounded to the nearest hundredth:
\[
\log_8 44.3 \approx 1.82
\]

#### 3) \( \log_5 63 \)
Using the Change of Base Formula:
\[
\log_5 63 = \frac{\ln 63}{\ln 5}
\]
Calculate:
\[
\ln 63 \approx 4.143, \quad \ln 5 \approx 1.609
\]
\[
\log_5 63 = \frac{4.143}{1.609} \approx 2.574
\]
Rounded to the nearest hundredth:
\[
\log_5 63 \approx 2.57
\]

#### 4) \( \log_8 23 \)
Using the Change of Base Formula:
\[
\log_8 23 = \frac{\ln 23}{\ln 8}
\]
Calculate:
\[
\ln 23 \approx 3.135, \quad \ln 8 \approx 2.079
\]
\[
\log_8 23 = \frac{3.135}{2.079} \approx 1.508
\]
Rounded to the nearest hundredth:
\[
\log_8 23 \approx 1.51
\]

#### 5) \( \log_8 14.2 \)
Using the Change of Base Formula:
\[
\log_8 14.2 = \frac{\ln 14.2}{\ln 8}
\]
Calculate:
\[
\ln 14.2 \approx 2.655, \quad \ln 8 \approx 2.079
\]
\[
\log_8 14.2 = \frac{2.655}{2.079} \approx 1.277
\]
Rounded to the nearest hundredth:
\[
\log_8 14.2 \approx 1.28
\]

#### 6) \( \log_8 12.9 \)
Using the Change of Base Formula:
\[
\log_8 12.9 = \frac{\ln 12.9}{\ln 8}
\]
Calculate:
\[
\ln 12.9 \approx 2.556, \quad \ln 8 \approx 2.079
\]
\[
\log_8 12.9 = \frac{2.556}{2.079} \approx 1.229
\]
Rounded to the nearest hundredth:
\[
\log_8 12.9 \approx 1.23
\]

#### 7) \( \log_8 185 \)
Using the Change of Base Formula:
\[
\log_8 185 = \frac{\ln 185}{\ln 8}
\]
Calculate:
\[
\ln 185 \approx 5.220, \quad \ln 8 \approx 2.079
\]
\[
\log_8 185 = \frac{5.220}{2.079} \approx 2.509
\]
Rounded to the nearest hundredth:
\[
\log_8 185 \approx 2.51
\]

#### 8) \( \log_{16} 17.2 \)
Using the Change of Base Formula:
\[
\log_{16} 17.2 = \frac{\ln 17.2}{\ln 16}
\]
Calculate:
\[
\ln 17.2 \approx 2.844, \quad \ln 16 \approx 2.773
\]
\[
\log_{16} 17.2 = \frac{2.844}{2.773} \approx 1.026
\]
Rounded to the nearest hundredth:
\[
\log_{16} 17.2 \approx 1.03
\]

#### 9) \( \log_5 12.8 \)
Using the Change of Base Formula:
\[
\log_5 12.8 = \frac{\ln 12.8}{\ln 5}
\]
Calculate:
\[
\ln 12.8 \approx 2.550, \quad \ln 5 \approx 1.609
\]
\[
\log_5 12.8 = \frac{2.550}{1.609} \approx 1.585
\]
Rounded to the nearest hundredth:
\[
\log_5 12.8 \approx 1.59
\]

#### 10) \( \log_{12} 108 \)
Using the Change of Base Formula:
\[
\log_{12} 108 = \frac{\ln 108}{\ln 12}
\]
Calculate:
\[
\ln 108 \approx 4.682, \quad \ln 12 \approx 2.485
\]
\[
\log_{12} 108 = \frac{4.682}{2.485} \approx 1.884
\]
Rounded to the nearest hundredth:
\[
\log_{12} 108 \approx 1.88
\]

#### 11) \( \log 5 \)
This is a common logarithm (\( \log_{10} 5 \)):
\[
\log 5 \approx 0.699
\]

#### 12) \( \log 4.4 \)
This is a common logarithm (\( \log_{10} 4.4 \)):
\[
\log 4.4 \approx 0.643
\]

#### 13) \( \log_3 34.2 \)
Using the Change of Base Formula:
\[
\log_3 34.2 = \frac{\ln 34.2}{\ln 3}
\]
Calculate:
\[
\ln 34.2 \approx 3.530, \quad \ln 3 \approx 1.099
\]
\[
\log_3 34.2 = \frac{3.530}{1.099} \approx 3.212
\]
Rounded to the nearest hundredth:
\[
\log_3 34.2 \approx 3.21
\]

#### 14) \( \log_8 37.2 \)
Using the Change of Base Formula:
\[
\log_8 37.2 = \frac{\ln 37.2}{\ln 8}
\]
Calculate:
\[
\ln 37.2 \approx 3.616, \quad \ln 8 \approx 2.079
\]
\[
\log_8 37.2 = \frac{3.616}{2.079} \approx 1.739
\]
Rounded to the nearest hundredth:
\[
\log_8 37.2 \approx 1.74
\]

#### 15) \( \log_{16} 34.6 \)
Using the Change of Base Formula:
\[
\log_{16} 34.6 = \frac{\ln 34.6}{\ln 16}
\]
Calculate:
\[
\ln 34.6 \approx 3.542, \quad \ln 16 \approx 2.773
\]
\[
\log_{16} 34.6 = \frac{3.542}{2.773} \approx 1.277
\]
Rounded to the nearest hundredth:
\[
\log_{16} 34.6 \approx 1.28
\]

#### 16) \( \log_{11} 161 \)
Using the Change of Base Formula:
\[
\log_{11} 161 = \frac{\ln 161}{\ln 11}
\]
Calculate:
\[
\ln 161 \approx 5.081, \quad \ln 11 \approx 2.398
\]
\[
\log_{11} 161 = \frac{5.081}{2.398} \approx 2.120
\]
Rounded to the nearest hundredth:
\[
\log_{11} 161 \approx 2.12
\]

Final Answers:

\[
\boxed{
\begin{array}{ll}
1) & 4.93 \\
2) & 1.82 \\
3) & 2.57 \\
4) & 1.51 \\
5) & 1.28 \\
6) & 1.23 \\
7) & 2.51 \\
8) & 1.03 \\
9) & 1.59 \\
10) & 1.88 \\
11) & 0.699 \\
12) & 0.643 \\
13) & 3.21 \\
14) & 1.74 \\
15) & 1.28 \\
16) & 2.12 \\
\end{array}
}
\]