How do you write these expressions as single logarithms? 1. #4 log 2 + log 6# 2. #3log_2 6-2# 3. #5 log 3 - 2 log 8#
2 Answers
Feb 17, 2018
#log 96# #log_2 54# #log (243/64)#
Explanation:
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#4 log 2 + log 6#
to start off, write the expression using exponents.
#log 2^4 + log 6#
since we are adding, we multiply
#log(2^4*6)#
#=log(16*6)#
#=log 96# -
#3 log_2 6 - 2#
for this one, we need to make sure the bases are the same so that we are able to combine them.
#log_2 6^3 - log_2 2^2#
since are subtracting, we divide
#log_2(6^3/2^2)#
#=log_2(216/4)#
#=log_2 54# -
#5 log 3 - 2 log 8#
#log 3^5 - log 8^2#
#log(3^5/8^2)#
#=log(243/64)#
Feb 17, 2018
#log(96)# - Refer to answer below
#log(243/64)#
Explanation:
-
When adding two logs together you multiply
#log(2)+log(6)#
#log(2*6)#
If there is a number behind the log it becomes an exponent
#4log(2)+log(6)#
#log(2^4*6)#
#log(96)# -
Refer to answer below
-
When you subtract logs you divide
#5log(3)−2log(8)#
#log(3^5/2^8)#
#log(243/64)#
