How do you simplify #Log(100.0/5.7)#?

1 Answer
Jun 16, 2017

Answer:

#log(100/5.7)=color(magenta)(2-log(5.7))#

Explanation:

(I am assuming the standard usage: #log# means #log_10#)

#log(a/b)=log(a)-log(b)#

#log_10(100)=color(blue)2color(white)("XXXX")#since #10^color(blue)2=100#

Therefore
#color(white)("XXX")log_10(100/5.7)=2 - log_10(5.7)#

Note that there is no simple evaluation for #log_10(5.7)#
but if necessary you could use a calculator to determine
#color(white)("XXX")log(5.7)~~0.755874856#
and from there
#color(white)("XXX")log_10(100/5.7)=2-log_10(5.7)~~1.244125144#