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

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(B) Given the equation $x^{\log_x(1-x)^2} = 9$.Using the property $x^{\log_x(A)} = A$,we have $(1-x)^2 = 9$.Expanding the equation: $1 - 2x + x^2 = 9$

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$\{4\}$

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(B) Given the equation $x^{\log_x(1-x)^2} = 9$.Using the property $x^{\log_x(A)} = A$,we have $(1-x)^2 = 9$.Expanding the equation: $1 - 2x + x^2 = 9$.Rearranging terms: $x^2 - 2x - 8 = 0$.Factoring the quadratic: $(x - 4)(x + 2) = 0$.This gives $x = 4$ or $x = -2$.However,for the term $\log_x(1-x)^2$ to be defined,the base $x$ must satisfy $x > 0$ and $x 
eq 1$.Checking $x = 4$: Base is $4 > 0$ and $4 
eq 1$. Also,$(1-4)^2 = 9 > 0$. This is a valid solution.Checking $x = -2$: Base is $-2$,which is not allowed $(x > 0)$.Thus,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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