How do you solve #log_10 18 - log_10 3x = log_10 2#?

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Answer 1 · 2016-10-13T21:23:53

#x=3#

#log_10 18 - log_10 3x = log_10 2#

Two log terms can be condensed into one.....
If logs re being subtracted, the numbers are being divided.

#log (18/(3x)) = log 2" "larr" " log A = logB rarr A=B#

#:. 18/(3x) = 2" "larr # now cross -multiply

#18 = 6x#

#3 = x#

#x = 3#

Answer 2 · 2016-10-13T21:45:35

#x=3#

#log(18)-log(3x)=log(2)#

Using the logarithmic quotient rule #-# #log(x) - log(y)=log(x/y)# #-# this equation can be rewritten as:

#log(18/(3x))=log(2)#

If we raise each term of the equation to #10#, we can remove the #log#s and set the terms equal to each other. Proof of why this can be done.

#10^log(18/(3x))=10^log(2)#

#18/(3x)=2#

#18=6x#

#x=18/6=3#