Solve using factorization method #log_10(3x²+8)=1+log_10 (x/2+1)#?

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Answer 1 · 2018-05-01T06:17:54

#log_10(3x²+8)= 1+log_10 (x/2+1)#

#=>log_10(3x²+8)=log_10 10+log_10 (x/2+1)#

#=>log_10(3x²+8)=log_10 (10*(x/2+1))#

#=>log_10(3x²+8)-log_10 (5x+10)=0#

#=>log_10[(3x²+8)/(5x+10)]=0#

#=>[(3x²+8)/(5x+10)]=10^0=1#

#=>(3x²+8)=(5x+10)#

#=>3x²+8-5x-10=0#

#=>3x²-5x-2=0#

#=>3x²-6x+x-2=0#

#=>3x(x-2)+1(x-2)=0#

#=>(x-2)(3x+1)=0#

#=>x=2orx=-1/3#