The solution set of the equation x^{log_x(1 - | vedclass

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(B) Given the equation: $x^{\log_x(1 - x)^2} = 9$.Using the logarithmic property $a^{\log_a(M)} = M$,the equation simplifies to:$(1 - x)^2 = 9$.Expand

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$\{4\}$

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(B) Given the equation: $x^{\log_x(1 - x)^2} = 9$.Using the logarithmic property $a^{\log_a(M)} = M$,the equation simplifies to:$(1 - x)^2 = 9$.Expanding the square:$1 - 2x + x^2 = 9$.Rearranging the terms to form a quadratic equation:$x^2 - 2x - 8 = 0$.Factoring the quadratic equation:$(x - 4)(x + 2) = 0$.This gives the potential solutions $x = 4$ and $x = -2$.Now,we must check the validity of these solutions in the original logarithmic expression $\log_x(1 - x)^2$. The base of a logarithm $x$ must satisfy $x > 0$ and $x 
eq 1$. For $x = 4$: The base is $4$,which is valid ($4 > 0$ and $4 
eq 1$).For $x = -2$: The base is $-2$,which is invalid as the base of a logarithm must be positive.Therefore,the only valid solution is $x = 4$.

The solution set of the equation $x^{\log_x(1 - x)^2} = 9$ is

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