How do you condense #Log_4 (20) - Log_4 (45) + log_4 (144)#?

2 Answers
May 27, 2018

Answer:

#3#

Explanation:

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx+logy=log(xy)#

#•color(white)(x)logx-logy=log(x/y)#

#•color(white)(x)log_b x=nhArrx=b^n#

#log_4 20+log_4 144-log_4 45#

#=log_4((20xx144)/45)#

#=log_4(64)=n#

#64=4^3=4^nrArrn=3#

May 27, 2018

Answer:

#log_4(20/45*144) = log_4(64) = 3#

Explanation:

The log product and quotient rules allow us to combine these terms, as long as the logs have the same base:
#log_b(x) + log_b(y) = log_b(x*y)#
#log_b(x) - log_b(y) = log_b(x/y)#

All the terms in this problem have logs of base 4, so we can apply these rules:
#log_4(20) - log_4(45) + log_4(144)#
# = log_4(20/45*144)#
# = log_4(64)#

Of course, the result of a log is the exponent of the base that will give you that number: #log_b (x) = y# means that #b^y = x#
So, since #4^3 = 64#, then
#log_4(64) = 3#