If (x^2 log _x 27) cdot log _9 x = x + 4, the | vedclass

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(A) Given equation: $(x^2 \log _x 27) \cdot \log _9 x = x + 4$Using the property $\log _a b \cdot \log _b c = \log _a c$, we have $\log _x 27 \cdot \l

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$2$

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(A) Given equation: $(x^2 \log _x 27) \cdot \log _9 x = x + 4$Using the property $\log _a b \cdot \log _b c = \log _a c$, we have $\log _x 27 \cdot \log _9 x = \log _9 27$.So, the equation becomes $x^2 \cdot \log _9 27 = x + 4$.Since $\log _9 27 = \log _{3^2} 3^3 = \frac{3}{2} \log _3 3 = \frac{3}{2}$, the equation is $x^2 \cdot \frac{3}{2} = x + 4$.Multiplying by $2$, we get $3x^2 = 2x + 8$, which simplifies to $3x^2 - 2x - 8 = 0$.Factoring the quadratic: $3x^2 - 6x + 4x - 8 = 0 \Rightarrow 3x(x - 2) + 4(x - 2) = 0$.This gives $(3x + 4)(x - 2) = 0$, so $x = 2$ or $x = -\frac{4}{3}$.Since the base of a logarithm $x$ must be positive and $x 
eq 1$, we reject $x = -\frac{4}{3}$.Therefore, $x = 2$.

If $(x^2 \log _x 27) \cdot \log _9 x = x + 4$, then the value of $x$ is

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