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April 18, 2015

April 18, 2015

Posted by **Ana** on Friday, May 28, 2010 at 10:26pm.

Use log(base12)3= 0.4421 and log(base12)7= 0.7831.

log(base12)36 = log(base12)3^2*4= 4.8842

log(base12)27/49 = -0.2399

log(base12)81/49 = 0.2022

log(base12)16,807 = 3.9155 (16,807= 7^5)

log(base12)441 = 2.4504 (441= 7^2*3^2)

- Chemistry/Algebra 2 -
**bobpursley**, Friday, May 28, 2010 at 10:56pmThe first is wrong, but a very large margin. 12^4= much more than 36

the others are correct.

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