How do you solve #Log (x +3)(x-2) = Log (7x-11)#?

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Answer 1 · 2015-11-17T10:47:15

This equation has one solution #x=5#

First we have to find the domain of this equation. Since #log# functions are only defined for positive real numbers we have to solve inequalities:

#(x+3)(x-2)>0# and #7x-11>0#

From first inequality we get: #x<-3 vv x>2#

From the second we get: #x> 1 4/7#

So we can write, that the domain is: #D=(2;+oo)#

Now we can solve the equation:

#log(x+3)(x-2)=log(7x-11)#

Since the base of logarythm is the same on both sides (none is written, so I suppose it is #10#), we can skip #log# signs and solve equation:

#(x+3)(x-2)=7x-11#

#x^2+x-6=7x-11#

#x^2-6x+5=0#

#Delta=6^2-4*1*5#

#Delta=36-20=16#

#sqrt(Delta)=4#

#x_1=(-b-sqrt(Delta))/(2a)#

#x_1=(6-4)/2=1#

#x_2=(-b+sqrt(Delta))/(2a)#

#x_2=(6+4)/2#

#x_2=5#

Since #x_1 !in D#, the only solution of this equation is #x_2=5#