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  • What is the solution of the inequation  <img alt="\\\mathrm{\log_{x-3}(2(x^2-10x+24))\geq\log_{(x-3)}(x^2-9)}" src="https://learn.careers360.com/latex-image/?%5C%

What is the solution of the inequation 

\\\mathrm{\log_{x-3}(2(x^2-10x+24))\geq\log_{(x-3)}(x^2-9)}

Option: 1

x\in[4, \infty)


Option: 2

x\in(-\infty,-3)\cup [10+\sqrt{43}, \infty)


Option: 3

x\in[10-\sqrt{43}, \infty)


Option: 4

x\in[10+\sqrt{43}, \infty)


Answers (1)

best_answer

 

 

Logarithmic Equations -

\\\mathrm{Equation \;of \;the \;form \; log_{\; a}f(x) = b \; (a>0), a \neq 1,} \\\mathrm{is\;known \;as\; logarithmic\; equation.} \\\mathrm{this\; is\; equivalent\; to\; the \; equation\; f(x)=a^b\; (f(x)>0) }

 

If the given equation is in the form of f (loga x) = 0, where a > 0 and a is not equal to 1.

In this case, put loga x = t and solve f(t) = 0.

 

-

this inequation is equivalent to:

 

\\\mathrm{(2(x^2-10x+24))\geq(x^2-9)} \\\mathrm{x^2-9>0} \\\mathrm{x-3>1} \\\mathrm{on\;solving\; these\; equation\; we \; get} \\\mathrm{x\in[10+\sqrt{43}, \infty)}

 

option (d) is correct

Posted by

Gautam harsolia

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