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The set of values of b for which $2\log_{1/36}(bx+28)=-\log_6(12-4x-x^2)$ has exactky one solution is

$\begin{array}{1 1}(A)\;(-\large\frac{14}{9},\normalsize\infty)&(B)\;(-\infty,-14)\cup (\large\frac{14}{3},\normalsize \infty)\\(C)\;(-\infty,-14)\cup \{4\}\cup (\large\frac{14}{3},\normalsize \infty)&(D)\;\{4\}\end{array} $

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$\Rightarrow \log_{1/16}(bx+28)=\log_{1/6}(12-4x-x^2)$
$\Rightarrow bx+28=12-4x-x^2$
$\Rightarrow x^2+(b+4)x+16=0$
For unique soluion $D=0$
$\Rightarrow (b+4)^2-4(1)(16)=0$
$\Rightarrow (b+4)^2-64=0$
$\Rightarrow b=4,-12$
$b=-12$ is not possible
So $b=4$ is only possible value.
Hence (D) is the correct answer.
answered Apr 15, 2014 by sreemathi.v

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