The number of solutions of the equation log_{ | vedclass

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Question

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

Vedclass · Accepted Answer

$0$

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(A) Given the equation: $\log_{\sqrt{3}}(x^{3} - 1) = \log_{\sqrt{3}}(x - 1) + 2$For the logarithmic terms to be defined,we must have $x^{3} - 1 > 0$ and $x - 1 > 0$,which implies $x > 1$.Using the property $\log_{b}(a) - \log_{b}(c) = \log_{b}(\frac{a}{c})$,we get:$\log_{\sqrt{3}}(\frac{x^{3} - 1}{x - 1}) = 2$Since $x^{3} - 1 = (x - 1)(x^{2} + x + 1)$,we have:$\log_{\sqrt{3}}(x^{2} + x + 1) = 2$Converting to exponential form:$x^{2} + x + 1 = (\sqrt{3})^{2}$$x^{2} + x + 1 = 3$$x^{2} + x - 2 = 0$Factoring the quadratic equation:$(x + 2)(x - 1) = 0$$x = -2$ or $x = 1$Since the domain requires $x > 1$,both $x = -2$ and $x = 1$ are rejected.Thus,the number of solutions is $0$.

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

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