How do you express a sum of logarithms for #log_5(125*25)#?

2 Answers
Aug 7, 2018

Answer:

#log_5(125xx25) = log_5 125 + log_5 25#

Explanation:

#log_b (x xx y) = log_b x + log_b y#

#log_5(125xx25) = log_5 125 + log_5 25 #

#"As a sum of logarithms: " log_5(125xx25) = log_5 125 + log_5 25#

Aug 7, 2018

Answer:

In support of mazoo's answer

Explanation:

#color(blue)("Preliminaries")#

Note that #log_a(x) = log_10(x)/log_10(a)#

Actually this still works if #a=10# as we have:

#log_10(x)=log_10(x)/log_10(10) = log_10(x)/1#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Looking at the question")#

Known: #125xx25=3125#

So we have: #log_5(3125) #

Converting to log base 10 this gives:

#log_10(3125)/log_10(5) =(3.49485...) /(0.69897...)#

#log_10(3125)/log_10(5) =5" "............Solution(1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("looking at mizoo's answer")#

They state the answer as #log_5(125)+log_5(25)#

#log_10(125)/log_10(5)+log_10(25)/log_10(5) #

#color(white)("ddd")3 color(white)("ddddd")+ color(white)("dd")2color(white)("dddd") = 5" "...Solution(2)#

As both solutions 1 and 2 match it demonstrates that mazoo has shown the correct method