Question

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.

• A. $(-\infty ,-14)\cup \left\{4\right\}\cup \left[\frac{14}{3},\infty \right)$
• B. $(-\infty ,-4)\cup \left[\frac{14}{3},\infty \right)$
• C. $(-\infty ,-14)\cup \left[\frac{14}{3},\infty \right)$
• D. $(-\infty ,-14)\cup \left\{4\right\}\cup \left[14,\infty \right)$

Answer 1

Answer: (A)

$(-\infty ,-14)\cup \left\{4\right\}\cup \left[\frac{14}{3},\infty \right)$

$2{\log }_{1/25}(bx+28)={\log }_{1/5}(12-4x-x^2)$

$\Rightarrow bx+28=12-4x-x^2$
$\Rightarrow x^2+(b+4)x+16=0$
For only one solution,
Case-I
$D=0$ or expression is perfect square $\Rightarrow b=4,-12$
At $b=4\Rightarrow x=-4$ satisfies both log domain.
At $b=-12\Rightarrow x=4$ does not satisfy the log domain.
$\Rightarrow b=4$
Domain $12-4x-x^2>0\Rightarrow x^2+4x-12<0$
$\Rightarrow -6<x<2\&bx+28>0$
Case-II
Only one root in domain of log.
$\Rightarrow f(-6)\cdot f\left(2\right)<0$
$\Rightarrow (28-6b)(28+2b)<0$
$\Rightarrow b\in (-\infty ,14)\cup (\frac{14}{3},\infty )$
At $x=-6$ at $b=\frac{14}{3},x$ will still satisfy the domain.
Similarly, at $x=2\Rightarrow b=-14$
$x$ does not satisfy the domain.
$\Rightarrow b\in (-\infty ,-14)\cup \left\{4\right\}\cup \left[\frac{14}{3},\infty \right)$