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

Related questions
See all questions in Common Logs
Impact of this question
1755 views around the world
You can reuse this answer
Creative Commons License