- #1
John O' Meara
- 330
- 0
I wish to find the value of .8^(2/5) using logs. I can find the value of .8^(-2/5) as follows: =(log(8)X1/10)X-2/5
=(-1 + .9031)X-2/5 = (-.0969)X-2/5 = +.03876;
antilog(.03876) = 1.093;
Now to find .8^(2/5) my approach is the same:
log(.8)X2/5 = (log(8)X10^-1)X2/5
= (bar1 + .9031)X2/5 : what do I do next. (bar1 = -1)
=(-1 + .9031)X-2/5 = (-.0969)X-2/5 = +.03876;
antilog(.03876) = 1.093;
Now to find .8^(2/5) my approach is the same:
log(.8)X2/5 = (log(8)X10^-1)X2/5
= (bar1 + .9031)X2/5 : what do I do next. (bar1 = -1)