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Solution of \log _{(0.2)}(x+2) > \log _{(0.2)}(2 x) is

Option: 1

\left ( 2,\infty \right )

 

 

 


Option: 2

\left ( -\infty,\infty \right )


Option: 3

\left ( -2,\infty \right )


Option: 4

\left ( 0,\infty \right )


Answers (1)

best_answer

\begin{aligned} & \log _{(0.2)}(x+2)>\log _{(0.2)}(2 x) \\ \end{aligned}

\begin{aligned} \Rightarrow &(x+2) <2 x \end{aligned}                (As base <1)

\begin{aligned} \Rightarrow x> 2 \end{aligned}

Domains

\begin{aligned} \cdot \: \: x+2 &>0 \Rightarrow x>-2 \\ \end{aligned}

\begin{aligned} \cdot \: \: 2 x &>0 \Rightarrow x>0 \end{aligned}

Intersection

 

x> 2

 

 

Posted by

sudhir.kumar

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