Let the domain of the function $f(x)=\log_3\log_5\log_7(9x-x^2-13)$ be the interval $(m,n)$. Let the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ have eccentricity $\dfrac{n}{3}$ and the length of the latus rectum $\dfrac{8m}{3}$. Then $b^2-a^2$ is equal to:
Step-by-Step Solution
Key Concept: $\log_7(\cdot)>0\Rightarrow9x-x^2-13>1\Rightarrow(x-2)(x-7)<0\Rightarrow x\in(2,7)$. $\log_5(\cdot)>0\Rightarrow\log_7(\cdot)>1\Rightarrow9x-x^2-13>7\Rightarrow(x-4)(x-5)<0\Rightarrow x\in(4,5)$. Domain $=(4,5)$, so $m=4$, $n=5$.
$m=4$, $n=5$, $a=3$, $b^2=16$. $b^2-a^2=7$.
Correct Answer: 1