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
  1. #log 96#
  2. #log_2 54#
  3. #log (243/64)#

Explanation:

  1. #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#

  2. #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#

  3. #5 log 3 - 2 log 8#
    #log 3^5 - log 8^2#
    #log(3^5/8^2)#
    #=log(243/64)#

Feb 17, 2018
  1. #log(96)#
  2. Refer to answer below
  3. #log(243/64)#

Explanation:

  1. 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)#

  2. Refer to answer below

  3. When you subtract logs you divide
    #5log(3)−2log(8)#
    #log(3^5/2^8)#
    #log(243/64)#