How do you solve #log_4(v+8)=log_4(-4v-2)#?

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Answer 1 · 2016-09-26T10:29:44

#v=-2#

#log_4(v+8)=log_4(-4v-2)#

#hArrlog_4(v+8)-log_4(-4v-2)=0#

or #log_4((v+8)/(-4v-2))=0#

Therefore #((v+8)/(-4v-2))=4^0=1# or

#v+8=-4v-2# or

#v+4v=-2-8# or

#5v=-10# or #v=-2#