How do you condense #log4+ log2-log5 #?

1 Answer
Jun 18, 2016

#log(8/5)#

Explanation:

Using the following #color(blue)"laws of logarithms"#

#•logx+logyhArrlog(xy)........(1)#

#•logx-logyhArrlog(x/y)........(2)#

using (1) : #log4+log2=log(4xx2)=log8#

using (2) : #log8-log5=log(8/5)#

#rArrlog4+log2-log5=log(8/5)#