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Mathematics 32 Online
OpenStudy (anonymous):

how do you approximate the value of log 72?

OpenStudy (anonymous):

please anyone!

OpenStudy (anonymous):

anything?

OpenStudy (anonymous):

oh my

OpenStudy (anonymous):

Just for clarification, is this the logarithm specifically with base of 10?

OpenStudy (anonymous):

yes

OpenStudy (perl):

factor 72 ?

OpenStudy (anonymous):

and then what?

OpenStudy (anonymous):

\[x = \log_{10} 72\] \[72 = 10^{x}\]

OpenStudy (agent0smith):

I'm assuming you mean without a calculator... You can do it by trial and error. You know that log72 will be somewhere between 1 and 2, since log10 = 1 and log100 = 2. It won't be easy w/o a calculator, really. This table will help you estimate: http://www.rapidtables.com/math/algebra/logarithm/Logarithm_Table.htm You could also factor 72 and break it up into a couple of logarithms, that isn't going to make it much easier, though. Since 72 = 9x8, log72 = log(9x8) = log9 + log8. From here you can get ~exactly the values you need (just find 9.00 and 8.00 and add the given values together): http://www.sosmath.com/tables/logtable/logtable.html

OpenStudy (agent0smith):

From the table, log8.00 is 0.9030900 log9.00 is 0.9542425 log72 = log9 + log8 = 0.9030900 + 0.9542425

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