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Find the values of 'b' for which the equation

$$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$ has only one solution. =$$-2/2\log_{5}(bx+28)=-\log_5(12-4x-x^2)$$ My try:

After removing the logarithmic terms I get the quadratic $x^2+x(b+4)+16=0$ Putting discriminant equal to $0$ I get $b={4,-12}$ But $-12$ cannot be a solution as it makes $12-4x-x^2$ negative so I get $b=4$ as the only solution.

But the answer given is $(-\infty,-14]\cup{4}\cup[14/3,\infty)$.I've no idea how.Help me please.

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  • $\begingroup$ You can't remove the logarithmic term like that, the bases aren't the same. $\endgroup$
    – Zain Patel
    Jul 17, 2016 at 11:04
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    $\begingroup$ I adjusted the bases before removing @ZainPatel.Check. $\endgroup$
    – user220382
    Jul 17, 2016 at 11:04
  • $\begingroup$ Because there is a root that should be ignored. $\endgroup$
    – Zack Ni
    Jul 17, 2016 at 11:15
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    $\begingroup$ Which root? @ZackNi $\endgroup$
    – user220382
    Jul 17, 2016 at 11:18
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    $\begingroup$ I have read it. It shows that you only consider the discriminant equals to 0 but not the case that there are two real roots and one should be ignored. $\endgroup$
    – Zack Ni
    Jul 17, 2016 at 11:23

1 Answer 1

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You have $$x^2+(4+b)x+16=0\tag1$$ This is correct.

However, note that when we solve $$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$ we have to have $$bx+28\gt 0\quad\text{and}\quad 12-4x-x^2\gt 0,$$ i.e. $$bx\gt -28\quad\text{and}\quad -6\lt x\lt 2\tag2$$

Now, from $(1)$, we have to have $(4+b)^2-4\cdot 16\geqslant 0\iff b\leqslant -12\quad\text{or}\quad b\geqslant 4$.

Case 1 : $b\lt -14$

$$(2)\iff -6\lt x\lt -\frac{28}{b}$$

Let $f(x)=x^2+(4+b)x+16$. Then, since the equation has only one solution, we have to have $$f(-6)f\left(-\frac{28}{b}\right)\lt 0\iff b\lt -14$$ So, in this case, $b\lt -14$.

Case 2 : $-14\leqslant b\leqslant -12$ or $4\leqslant b\lt \frac{14}{3}$

$$(2)\iff -6\lt x\lt 2$$

$b=4$ is sufficient, and $b=-12$ is not sufficient. For $b\not=4,-12$, $$f(-6)f(2)\lt 0\iff b\lt -14\quad\text{or}\quad b\gt \frac{14}{3}$$ So, in this case, $b=4$.

Case 3 : $b\geqslant \frac{14}{3}$

$$(2)\iff -\frac{28}{b}\lt x\lt 2$$ $b=\frac{14}{3}$ is sufficient. For $b\gt\frac{14}{3}$, $$f\left(-\frac{28}{b}\right)f(2)\lt 0\iff b\gt \frac{14}{3}$$ So, in this case, $b\geqslant 14/3$.

Therefore, the answer is $$\color{red}{(-\infty,-14)\cup{4}\cup\bigg[\frac{14}{3},\infty\bigg)}$$

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  • $\begingroup$ Don't have to do so much.Just f(-6)f(2)<0 ensures there is exactly one real root between the valid bound's of x... $\endgroup$
    – user220382
    Jul 17, 2016 at 11:57
  • $\begingroup$ @ZOZ: $f(-6)f(2)\lt 0$ enables us to eliminate the case where there are two real roots, but if we only consider $f(-6)f(2)\lt 0$, it is possible that there is no real roots in the interval where $bx+28\gt 0$ and $-6\lt x\lt 2$. $\endgroup$
    – mathlove
    Jul 17, 2016 at 12:14
  • $\begingroup$ I get your comment.But I still have difficulty in understanding the cases into which you divided the problem.Like Why did you take the first case as b<-14 ? $\endgroup$
    – user220382
    Jul 17, 2016 at 12:16
  • $\begingroup$ @ZOZ: We have $-6, 2$ and $-28/b$ as the endpoints of the intervals we are interested in. Now, $-14$ comes from $b$ such that $2=-28/b$. Also, $14/3$ comes from $-6=-28/b$. I hope this helps. $\endgroup$
    – mathlove
    Jul 17, 2016 at 12:19
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    $\begingroup$ @mathophile : Thanks. You are right that $b=-14$ has to be excluded, but $b=\frac{14}{3}$ has to be included since $x=-\frac 83$ is the only solution. $\endgroup$
    – mathlove
    Jan 10 at 17:14

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