2
$\begingroup$

Given
$$\frac{1}{\log_{4}\left(\frac{x+1}{x+2}\right)}<\frac{1}{\log_{4}(x+3)}.$$ Then what is the range of values of $x$ for which this inequality is satisfied.

My Try On simplification, I got $x$ from $(-\infty,-2)$. That however is not the correct answer. Can someone tell me each and every step and every individual domains and inequalities I need to consider in this?

$\endgroup$
  • 1
    $\begingroup$ its true for all $x > -1$ since the left side will always be negative while the right side will always be positive. $\endgroup$ – Ahmad Aug 21 '17 at 15:13
1
$\begingroup$

The arguments of the logarithms should be positive. Hence $x+3>0$ and $\frac{x+1}{x+2}>0$, that is $x\in (-3,-2)\cup (-1,+\infty).$

If $x\in (-3,-2)$ then $0<x+3<1$ and $\frac{x+1}{x+2}>1$. Therefore $$\log_4(x+3)<0\quad\mbox{and}\quad\log_{4}\left(\frac{x+1}{x+2}\right)>0$$ and the inequality does not hold.

If $x\in (-1,+\infty)$ then then $x+3>1$ and $0<\frac{x+1}{x+2}<1$. Therefore $$\log_4(x+3)>0\quad\mbox{and}\quad\log_{4}\left(\frac{x+1}{x+2}\right)<0$$ and the inequality holds.

So the inequality holds iff $x>-1$. Note that if we replace the base $4$ with another number greater than $1$ then the result is the same.

$\endgroup$
  • $\begingroup$ Crisp, clear, perfect. Thanks. $\endgroup$ – Tanuj Aug 21 '17 at 15:53
0
$\begingroup$

HINT: you must do some case work: $$\log_4(x+3)>0$$ it is $$x+3>1$$ or $$x>-2$$ $$\log_4\left(\frac{x+1}{x+2}\right)>\log_41$$ this is equivalent to $$\frac{x+1}{x+2}>1$$ which is not true. can you solve the other cases? if $$\log_4(x+3)<\log_4 1$$ we get $$x<-2$$ if $$\log_4\left(\frac{x+1}{x+2}\right)>\log_41$$ we get $$\frac{x+1}{x+2}>1$$ and we get $$x+1<x+2$$ which is true. and you have to solve $$x+3>\frac{x+1}{x+2}$$ this is equivalent to $$(x+3)(x+2)<x+1$$ this gives $$x^2+4x+5<0$$ or $$(x+1)^2+1<0$$ which is impossible.

$\endgroup$
0
$\begingroup$

We can use the intervals method.

The domain gives $x>-1$ or $-3<x<-2$.

We need to solve that $$\frac{1}{\log_4(x+3)}-\frac{1}{\log_4\frac{x+1}{x+2}}>0$$ or $$\frac{\log_4\frac{x+1}{(x+2)(x+3)}}{\log_4(x+3)\log_4\frac{x+1}{x+2}}>0.$$ Now, since $\frac{x+1}{(x+2)(x+3)}\neq1$, it's enough to check two intervals only, which gives the answer: $$(-1,+\infty).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.