How do you use change of base to simplify #log_7 83#?

1 Answer
Gió
Feb 13, 2015

You use the fact that:
#log_aM=ln(M)/ln(a)#
In your case you have:
#log_7(83)=ln(83)/ln(7)=2.271#
That can be evaluated with a normal calculator.

You can now check your result writing:
#7^(2.271)=83.026..~~83#

(I use #ln#, the natural log, but you can use the other log that your calculator uses, the one in base 10, the result is the same, try it)

Hope it helps

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