The number of solutions of frac{log 5 + log ( | vedclass

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Question

(D) Given the equation: $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$For the expression to be defined,we must have $x - 2 > 0$ and $x - 2 
eq 1$

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given the equation: $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$For the expression to be defined,we must have $x - 2 > 0$ and $x - 2 
eq 1$,which implies $x > 2$ and $x 
eq 3$.Using the property $\log a + \log b = \log (ab)$,the equation becomes:$\log (5(x^2 + 1)) = 2 \log (x - 2)$Using the property $n \log a = \log (a^n)$,we get:$\log (5x^2 + 5) = \log ((x - 2)^2)$Equating the arguments:$5x^2 + 5 = x^2 - 4x + 4$$4x^2 + 4x + 1 = 0$$(2x + 1)^2 = 0$$x = -\frac{1}{2}$Since the domain condition is $x > 2$,the value $x = -\frac{1}{2}$ is not a valid solution.Therefore,the number of solutions is $0$.

The number of solutions of $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$ is

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