How do you solve #log ( x +1 ) + log 7 = log 14 - log ( 2-x ) #?

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Answer 1 · 2018-06-12T13:15:36

#x=1#

#log(x+1)+log7=log14-log(2-x)#

#log(x+1)+log(2-x)=log14-log7#

#log((x+1)/(2-x))=log2#

#(x+1)/(2-x)=2#

#x+1=4-2x#

#3x=3#

#x=1#

Recall:

#log_10 a + log_10 b = log_10(atimesb) = log_10 ab#

#log_10 a- log_10 b=log_10(adivide b)=log_10 (a/b)#