How do you evaluate #log_3 7# using the change of base formula?

1 Answer
Jan 7, 2017

Answer:

I found: #1.77124#

Explanation:

The change of base allows you to change from a base, say #b#, to a new base #c# as:
#log_b(x)=(log_c(x))/(log_c(b))#
Where the new base #c# can be choosen to be "easy" to evaluate; if you have a pocket calculator, the new base could be #e# that can be evaluated using the Natural Logarithm (#ln#) on the calculator.
We write:
#log_3(7)=(log_e(7))/(log_e(3))=(ln(7))/(ln(3))=1.77124#