Chemistry/Algebra 2
posted by Ana .
Are the answers to these correct?
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)

The first is wrong, but a very large margin. 12^4= much more than 36
the others are correct.
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