How do you solve # log_12 (x + 1) – log_12 4 = log_12 16#?

1 Answer
Gió
May 10, 2015

Here you need to remember some rules of logarithms!
Consider that:
#loga-logb=log(a/b)#
So you can write:
#log_12((x+1)/4)=log_12(16)#
Also remember that #a^(log_a(x))=x#
So use #12# to eliminat the logs as:
#12^(log_12((x+1)/4))=12^(log_12(16))#
which gives you:
#(x+1)/4=16#
And:
#x=63#

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