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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Answer

(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

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $7x^{2} + 16x - 15 = 0$
$II.$ $y^{2} - 6y - 7 = 0$

If $\alpha, \beta, \gamma$ are the roots of $x^3 - x - 2 = 0$,then the value of $\alpha^5 + \beta^5 + \gamma^5$ is-

If $x-\frac{1}{x}=\sqrt{21},$ then the value of $\left(x^{2}+\frac{1}{x^{2}}\right)\left(x+\frac{1}{x}\right)$ will be

If $\alpha, \beta$ are the roots of $(x - a)(x - b) = c,$ where $c \neq 0,$ then the roots of $(x - \alpha)(x - \beta) + c = 0$ are

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 3x + 4 = 0$,then the equation whose roots are $\alpha^2$ and $\beta^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module