Any solution of the inequality,$\frac{\log_3(x^2-3x+7)}{\log_3(3x+2)}<1$ is also a solution of the inequality,$x^2+(5-2a)x-10a\leq 0$

Question

Find out the values of $a$ for which any solution of the inequality,$\frac{\log_3(x^2-3x+7)}{\log_3(3x+2)}<1$ is also a solution of the inequality,$x^2+(5-2a)x-10a\leq 0$

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I first found out the domain of the inequality.
For that,$x^2-3x+7>0$ and $3x+2>0$ and $3x+2\ne 1$
Solving the above inequalities,we get
$x\in(\frac{-2}{3},\frac{-1}{3})\cup(\frac{-1}{3},\infty)$

$\log_3(3x+2)$ will be positive when $x>\frac{-1}{3}$ and $\log_3(3x+2)$ will be negative when $x\in(\frac{-2}{3},\frac{-1}{3})$.

Case $1:$When $x>\frac{-1}{3}$
$\frac{\log_3(x^2-3x+7)}{\log_3(3x+2)}<1$

$\log_3(x^2-3x+7)<\log_3(3x+2)\implies x^2-3x+7<3x+2$
$x^2-6x+5<0\implies (x-1)(x-5)<0\implies x\in(1,5)$

Case$2$:When $x\in(\frac{-2}{3},\frac{-1}{3})$
$\frac{\log_3(x^2-3x+7)}{\log_3(3x+2)}<1$

$\log_3(x^2-3x+7)>\log_3(3x+2)\implies x^2-3x+7>3x+2$
$x^2-6x+5>0\implies (x-1)(x-5)>0\implies x\in(-\infty,1)\cup(5,\infty)$

Combined with $x\in(\frac{-2}{3},\frac{-1}{3})$,we get
$x\in(\frac{-2}{3},\frac{-1}{3})$

Combining the case (1) and (2),we get $x\in(\frac{-2}{3},\frac{-1}{3})\cup(1,5)$

I took the second inequality $x^2+(5-2a)x-10a\leq 0$
So $(5-2a)^2+40a\geq 0$
$(5+2a)^2\geq 0$ which is always true.

I do not know how to find the values of $a$.The answer given is $a\geq \frac{5}{2}.$I do not know how they have found the values of $a$.

Answer 1

You have identified the set of possible solutions for $x$ that will arise from the log inequality, namely $$(-\frac 23,-\frac 13)\cup(1,5)$$

Now the second inequality is $$x^2+(5-2a)x-10a\leq0$$ $$\Rightarrow (x+5)(x-2a)\leq 0$$ $$\Rightarrow -5\leq x \leq 2a$$

This must be true for all $x$ in the solution set, so $$a\geq \frac 52$$

Answer 2

The trick here seems to be to observe that $$ \frac{\log_3(x^2-3x+7)}{\log_3(3x+2)}<1 $$ exactly if $\log_3 (x^2-3x+7) < \log_3 (3x+2)$, which in turn holds (because $\log_3$, like all logarithmns, is monotone!) exactly if $x^2-3x+7  < 3x+2$ or in other words if $$ x^2 -6x + 5 < 0. $$

Compare that to $x^2+(5-2a)x-10a < 0$, and you'll notice that the two inequalities are identical if you set $a=-\frac{1}{2}$. Since $x < y$ in particular implies $x \leq y$, you get that $$ \frac{\log_3(x^2-3x+7)}{\log_3(3x+2)}<1 \quad\Rightarrow\quad x^2+(5-2a)x-10a \leq 0 \quad \text{for}\quad a=-\frac{1}{2}. $$

I don't know how they (whoever that is) end up with $a \geq \frac{5}{2}$.

Answer 3

You have found the wrong range of x from the first equation. it is 1<=x<=5. And from there we get that 1,5 will both have to satisfy the second inequality, so a>=5/2