# 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

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.

• You can't remove the logarithmic term like that, the bases aren't the same. Jul 17, 2016 at 11:04
• I adjusted the bases before removing @ZainPatel.Check.
– user220382
Jul 17, 2016 at 11:04
• Because there is a root that should be ignored. Jul 17, 2016 at 11:15
• Which root? @ZackNi
– user220382
Jul 17, 2016 at 11:18
• 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. Jul 17, 2016 at 11:23

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

• 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...
– user220382
Jul 17, 2016 at 11:57
• @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 $-6\lt x\lt 2$. Jul 17, 2016 at 12:14
• 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 ?
– user220382
Jul 17, 2016 at 12:16
• @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. Jul 17, 2016 at 12:19
• @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. Jan 10 at 17:14