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Rewrite the following in terms of ln 2, ln 3, and ln 5. Then use ln 2=0.693, ln 3=1.099, and ln 5=1.609.
1. ln 6
2. ln (10/3)
3. ln 30
4. ln 12
5. ln (2/5)
5. ln (5/6)



Answer :

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[tex]\log_a (b \times c)=\log_a b + \log_a c \\ \log_a (\frac{b}{c})=\log_a b-\log_a c \\ \log_a b^c=c \log_a b \\ \\ 1. \\ \ln 6=\ln (2 \times 3)=\boxed{\ln 2 + \ln 3} \approx 0.693+1.099=\boxed{1.792} \\ \\ 2. \\ \ln (\frac{10}{3})=\ln (\frac{2 \times 5}{3})=\boxed{\ln 2 + \ln 5 - \ln 3} \approx 0.693+1.609-1.099=\boxed{1.203} \\ \\ 3. \\ \ln 30=\ln (2 \times 3 \times 5)=\boxed{\ln 2+ \ln 3 + \ln 5} \approx 0.693+ 1.099 + 1.609=\\=\boxed{3.401}[/tex]

[tex]4. \\ \ln 12=\ln (2^2 \times 3)=\ln 2^2 + \ln 3=\boxed{2 \ln 2+ \ln 3} \approx 2 \times 0.693+ 1.099= \\ =\boxed{2.485} \\ \\ 5. \\ \ln (\frac{2}{5})=\boxed{\ln 2 - \ln 5} \approx 0.693-1.609=\boxed{-0.916} \\ \\ 6. \\ \ln (\frac{5}{6})=\ln (\frac{5}{2 \times 3})=\ln 5-(\ln 2+ \ln 3)=\boxed{\ln 5 - \ln 2 - \ln 3} \approx \\ \approx1.609-0.693-1.099=\boxed{-0.183}[/tex]

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