How do you calculate # log_5(4) #?

1 Answer
Apr 8, 2016

Answer:

#log_5(4)=0.8614#

Explanation:

Let #log_ba=x#, then #b^x=a#.

If #a=10^n# and #b=10^m#, then #n=loga# and #m=logb# and

#b^x=a# becomes #(10^m)^x=10^n# or #10^(mx)=10^n# i.e.

#mx=n#

Hence #x=n/m=loga/logb#

Thus #log_5(4)=log4/log5=0.6021/0.6990=0.8614#