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 logarithmic property $a^{\log_a(b)} = b$,where $x > 0$ and $x 
eq 1$,the equation simplifie

Vedclass · Accepted Answer

$\{4\}$

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(B) Given the equation $x^{\log_x(1-x)^2} = 9$. Using the logarithmic property $a^{\log_a(b)} = b$,where $x > 0$ and $x 
eq 1$,the equation simplifies to: $(1-x)^2 = 9$. Expanding the square: $1 + x^2 - 2x = 9$. Rearranging into a standard quadratic form: $x^2 - 2x - 8 = 0$. Factoring the quadratic: $(x - 4)(x + 2) = 0$. This gives potential solutions $x = 4$ or $x = -2$. Checking the domain of the original expression $x^{\log_x(1-x)^2}$: The base of the logarithm $x$ must satisfy $x > 0$ and $x 
eq 1$. For $x = 4$,the base is $4$,which is valid. For $x = -2$,the base is $-2$,which is not valid 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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