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log(x+1) + log(x-5) = 4
USING THE FORMULA LOG A TO BASE X = LOG A/LOG X TO A COMMON BASE OR STANDARD BASE.HENCE WE HAVE
{LOG (X+1)/LOG 2} + {LOG (X-5)/LOG 2}=4
LOG(X+1)+LOG(X-5)=4*LOG 2=LOG 2^4=LOG 16 (SINCE LOG X^N=N*LOG X)
LOG(X+1)(X-5)=LOG 16 (SINCE LOG X + LOG Y =LOG(X*Y))
TAKING ANTILOGS
(X+1)(X-5)=16
X^2-5X+X-5=16
X^2-4X-5-16=0
X^2-7X+3X-21=0
X(X-7)+3(X-7)=0
(X-7)(X+3)=0
HENCE X=7 OR X=-3 AS X=-3 LEADS TO LOG OF NEGATIVE NUMBERS WHICH DO NOT EXIT , THE ANSWER IS X=7
