How do you evaluate #4log_4 6 - log_4 5#?

1 Answer
Gió
May 29, 2015

Use the following properties of logs:
1] #alogx=logx^a#
2] #logx-logy=log(x/y)#
So in your case:
#log_4(6^4)-log_4(5)=log_4(6^4/5)=log_4(259.2)=x#
so:
#4^x=259.2#
if #x=4#
#4^4=256#
if #x=4.002#
#4^(4.002)=259.2#

Another thing that you can do is to change base (using, for example, natural logs) as:
#log_4(259.2)=ln(259.2)/(ln(4))=4.0089#

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