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C3: intuitively knowing the values of nature logs and e

Hi everyone,

I'm going through a number of questions on natural logs and the number 4. For example:

eln92 e^{ \frac{ln9}{2}} and eln42 e^{ \frac{ln4}{2}}

The answers are nice whole numbers that have links to the original numbers in the logs (such as eln92 e^{ \frac{ln9}{2}} being the square root of 9, and eln42 e^{ \frac{ln4}{2}} being the square root of 4. I'm getting the answers by just punching the numbers in my calculator. Aside from familiarity, should I be recognising what these numbers are without using my calculator, or is the calculator the only way to go?
Reply 1
I guess you could be asked to find the value of, in which case you might need some working? Could be similar to how you could 'spot' the value in your head; can you do this?


Posted from TSR Mobile
Reply 2
It saves time in the exam just recognising them, e.g.

e^[ln(9)/2] is just the same as e^ln(9^1/2) which is e^ln(3), and since they're inverses of each other the answer is just 3.
(edited 10 years ago)
Reply 3
Original post by marcsaccount
Hi everyone,

I'm going through a number of questions on natural logs and the number 4. For example:

eln92 e^{ \frac{ln9}{2}} and eln42 e^{ \frac{ln4}{2}}

The answers are nice whole numbers that have links to the original numbers in the logs (such as eln92 e^{ \frac{ln9}{2}} being the square root of 9, and eln42 e^{ \frac{ln4}{2}} being the square root of 4. I'm getting the answers by just punching the numbers in my calculator. Aside from familiarity, should I be recognising what these numbers are without using my calculator, or is the calculator the only way to go?


You should be able to recognize square numbers from GCSE :smile:

9 = 3^2 so ln 9 = 2ln3 and (ln9)/2 = ln 3
4 = 2^2 so ln 4 = 2ln2 and (ln4)/2 = ln 2

You know that raising e to the power of something is the inverse operation to the natural logarithm, so you should be able to write down the answers straight away.
Reply 4
Davros, CTArsenal, cheers guys! I was getting a bit confused with the 2 in the denominator, which of course is just the same as multiply by 0.5 at the front and I can then use log laws. I just didn't see it at first. Thanks a lot. Have a nice evening!

Marc

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