Question

Solve for X, log(3.x2 + 12x: +19) - log(3x + 4) = 1

JEE/Engineering Exams

Maths

Solution

173

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( log left(3 x^{2}+12 x+19right)-log (3 x+4)=1 ) ( Rightarrow log left(frac{3 x^{2}+12 x+19}{3 x+4}right)=1 ) ( Rightarrow quad frac{3 x^{2}+12 x+19}{3 x+4}=10 ) ( Rightarrow 3 x^{2}+12 x+19=30 x+40 ) ( Rightarrow 3 x^{2}-18 x-21=0 ) ( Rightarrow x^{2}-6 x-7=0 Rightarrow x^{2}-7 x+x-7=0 ) ( Rightarrow(x-7)(x+1)=0 ) ( Rightarrow x=-1,7 )