ln Arithmetic

[Jason76]

Note: In below samples, C or D cannot equal \(\displaystyle 1\)

We are assuming the ln values have been evaluated so no absolute value signs.


Case A


\(\displaystyle A \ln (C) \pm A\ln (C) = \)


\(\displaystyle 5\ln(9) - 5 \ln(9) = \)


\(\displaystyle 5\ln(9) + 5 \ln(9) = \)


Case B:


\(\displaystyle A \ln (C) \pm B\ln (C) = \)


\(\displaystyle 5\ln(9) - 4 \ln(9) = \)


\(\displaystyle 5\ln(9) + 4 \ln(9) = \)




Case C:




\(\displaystyle A \ln (C) \pm A\ln (D) = \)


\(\displaystyle 5\ln(6) - 5 \ln(2) = \)


\(\displaystyle 5\ln(6) + 5 \ln(2) = \)


Case D:


\(\displaystyle A \ln (C) \pm B\ln (D) = \)


\(\displaystyle 5\ln(4) - 3 \ln(5) = \)


\(\displaystyle 5\ln(4) + 3 \ln(5) = \)

 

[Deleted member 4993]

Note: In below samples, C or D cannot equal \(\displaystyle 1\)


Case A


\(\displaystyle A \ln (C) \pm A\ln (C) = \)


\(\displaystyle 5\ln(9) - 5 \ln(9) = \)


\(\displaystyle 5\ln(9) + 5 \ln(9) = \)


Case B:


\(\displaystyle A \ln (C) \pm B\ln (C) = \)


\(\displaystyle 5\ln(9) - 4 \ln(9) = \)


\(\displaystyle 5\ln(9) + 4 \ln(9) = \)




Case C:




\(\displaystyle A \ln (C) \pm A\ln (D) = \)


\(\displaystyle 5\ln(6) - 5 \ln(2) = \)


\(\displaystyle 5\ln(6) + 5 \ln(2) = \)


Case D:


\(\displaystyle A \ln (C) \pm B\ln (D) = \)


\(\displaystyle 5\ln(4) - 3 \ln(5) = \)


\(\displaystyle 5\ln(4) + 3 \ln(5) = \)

What is it - that you are trying to convey?

Please post a question with a pithy description (explanation, things-to-do)!

 

[Steven G]

Note: In below samples, C or D cannot equal \(\displaystyle 1\)

We are assuming the ln values have been evaluated so no absolute value signs.


Case A


\(\displaystyle A \ln (C) \pm A\ln (C) = \)


\(\displaystyle 5\ln(9) - 5 \ln(9) = \)


\(\displaystyle 5\ln(9) + 5 \ln(9) = \)


Case B:


\(\displaystyle A \ln (C) \pm B\ln (C) = \)


\(\displaystyle 5\ln(9) - 4 \ln(9) = \)


\(\displaystyle 5\ln(9) + 4 \ln(9) = \)




Case C:




\(\displaystyle A \ln (C) \pm A\ln (D) = \)


\(\displaystyle 5\ln(6) - 5 \ln(2) = \)


\(\displaystyle 5\ln(6) + 5 \ln(2) = \)


Case D:


\(\displaystyle A \ln (C) \pm B\ln (D) = \)


\(\displaystyle 5\ln(4) - 3 \ln(5) = \)


\(\displaystyle 5\ln(4) + 3 \ln(5) = \)

Look I have been on this site for awhile now and have not gotten upset with anyone until now. Why do you not pose a question? Why do you not even make an attempt to solve the problems? My 10 year old daughter who never saw an ln in her life would know that 5ln(9)-5ln(9)=0 if she was told that 5ln(9) was just some strange looking number. Do you really not know this?! If the answer is that you do not know that when you subtract the same quantities that you get 0 then you do not belong in any math class whatsoever. If you do know that the result is 0 then why do you not state so?!! Did you read the url for this website? Do you know how to read?