The number of solutions of the equation log_{ | vedclass

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(A) Given equation: $\log_{\sqrt{3}}(x^3 - 1) = \log_{\sqrt{3}}(x - 1) + 2$.For the logarithmic functions to be defined,we must have $x^3 - 1 > 0$ and

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$0$

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(A) Given equation: $\log_{\sqrt{3}}(x^3 - 1) = \log_{\sqrt{3}}(x - 1) + 2$.For the logarithmic functions to be defined,we must have $x^3 - 1 > 0$ and $x - 1 > 0$,which implies $x > 1$.Using the property $\log_b(A) - \log_b(B) = \log_b(A/B)$,we get $\log_{\sqrt{3}}((x^3 - 1)/(x - 1)) = 2$.Since $x^3 - 1 = (x - 1)(x^2 + x + 1)$,the equation becomes $\log_{\sqrt{3}}(x^2 + x + 1) = 2$.Converting to exponential form: $x^2 + x + 1 = (\sqrt{3})^2 = 3$.So,$x^2 + x - 2 = 0$.Factoring the quadratic: $(x + 2)(x - 1) = 0$.This gives $x = -2$ or $x = 1$.Since the domain requires $x > 1$,both $x = -2$ and $x = 1$ are rejected.Thus,there are $0$ solutions.

The number of solutions of the equation $\log_{\sqrt{3}}(x^3 - 1) = \log_{\sqrt{3}}(x - 1) + 2$ is:

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