by Galaxy » Mon 11 April, 2011 23:07
we have Re: `X=(log(2.723)-21log(1.03)) / (log(1.07)-log(1.03))`
the log `logA^2=2*logA` which means you can bring the power to the front of the log.
`X=(log(2.723)-log(1.03^21)) / (log(1.07)-log(1.03))`
the rule of logarithm is if we have `log(a)-log(b)=log(a/b)` so we have:
`X=(log(2.723/1.03^21))/(log(1.07/1.03))=0.1654/0.01654=10`
`:.` The X=10.