Calculus: How to express ln (.16) in terms of ln 2 and ln 5
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Calculus: How to express ln (.16) in terms of ln 2 and ln 5

[From: ] [author: ] [Date: 11-05-11] [Hit: ]
Thanks in advance for any help-ln(0.2ln2-2ln5-first express 0.0.use the logarithmic rules: ln(ab)=lna+lnb,ln(0.16) = ln(2^2/5^2) = ln(2^2) - ln(5^2) = ln2 + ln2 - ln5 - ln5-.......
Express ln (.16) in terms of ln 2 and ln 5. I am NOT just looking for an answer, I need to know how to do the problem. I'll be quick to award BA. Thanks in advance for any help

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ln(0.16)=
ln(16/100)=
ln(2^4)-ln[(2^2)(5^2)]=
4ln2-(2ln2+2ln5)=
2ln2-2ln5

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first express 0.16 in terms of 2 and 5:
0.16 = 2^2/5^2
use the logarithmic rules: ln(ab)=lna+lnb, ln(a/b)=lna-lnb
ln(0.16) = ln(2^2/5^2) = ln(2^2) - ln(5^2) = ln2 + ln2 - ln5 - ln5

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.16 = (2/5)^2 2*2/5/5 Multiply means we add logs, dividing means we subtract logs.

2ln2 - 2ln5 is your answer
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