How do you solve #log_7x+log_7(x+5)=log_7(14)#?

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Answer 1 · 2016-12-08T21:22:07

#x= -7#

#x= +2#

Notice is that everything is to logs of the same base. This means that we can get rid of the logs.

Note that #log(a)+log(b)# is the same thing as #log(axxb)#

This gives us:

#log_7(x(x+5))=log_7(14)#

#=> x(x+5)=14#

#x^2+5x-14=0#
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Note that #2xx7=14" and "7-2=5#

#(x+7)(x-2)=0#

#x= -7#

#x= +2#