Fm queries
This query is : Resolved
20 January 2014
can any1 explain me hw to find value of ln 1.0375. ans is 0.036814.. and ln 1.066 0.0647..
I m nt geting hw to find this values...
20 January 2014
ln of 1.03 is .029559 & ln of 1.04 is .039221 from natural log tables.
Interpolating the above values for 1.0375 we get .0368055.
Similarly ln 1.066 can be found through ln1.06 & ln1.07
In case of .0647 we can find the value by writing .0647 as (6.47 / 10*10)i.e. ln.0647 = ln6.47-ln10-ln10.
We can find the value from Excel or natural log calculators also.
I hope you would be able to do it now. If not, please let me know.
Thanks!
Interpolating the above values for 1.0375 we get .0368055.
Similarly ln 1.066 can be found through ln1.06 & ln1.07
In case of .0647 we can find the value by writing .0647 as (6.47 / 10*10)i.e. ln.0647 = ln6.47-ln10-ln10.
We can find the value from Excel or natural log calculators also.
I hope you would be able to do it now. If not, please let me know.
Thanks!
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