How do you solve #log_2(-5x) = log_(2)3 + log_2(x+2)#?

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Answer 1 · 2016-01-15T17:17:16

#log_2(-5x)=log_2(3)+log_2(x+2)#

From #log# properties we know that:

#log_c(a*b)=log_c(a)+log_c(b)#

#implies log_2(-5x)=log_2{3(x+2)}#
#implies log_2(-5x)=log_2(3x+6)#

Also form #log# properties we know that:

If #log_c(d)=log_c(e)#, then #d=e#

#implies -5x=3x+6#
#implies 8x=-6#
#implies x=-3/4#