`X=(log(2.723)-21log(1.03)) / (log(1.07)-log(1.03))`

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`X=(log(2.723)-21log(1.03)) / (log(1.07)-log(1.03))`

Postby barnamah » Mon 11 April, 2011 22:51

logarithm question `X=(log(2.723)-21log(1.03)) / (log(1.07)-log(1.03))`
Explore and know. That is asked.
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Re: `X=(log(2.723)-21log(1.03)) / (log(1.07)-log(1.03))`

Postby 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.
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