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

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

Vedclass · Accepted Answer

None of these

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(D) Given equation: $\frac{\log 5 + \log (x^2 + 1)}{\log (x - 2)} = 2$.For the logarithmic terms 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: $5x^2 + 5 = x^2 - 4x + 4$.Rearranging terms: $4x^2 + 4x + 1 = 0$.This is a perfect square: $(2x + 1)^2 = 0$,which gives $x = -\frac{1}{2}$.However,the domain condition requires $x > 2$. Since $x = -\frac{1}{2}$ does not satisfy $x > 2$,there is no valid solution.Thus,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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