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
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
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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2ln2 - 2ln5 is your answer
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