Let the domain of the function f(x) = log_{3} | vedclass

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(D) For the function $f(x) = \log_{3}\log_{5}\log_{7}(9x - x^{2} - 13)$ to be defined, we require: $\log_{5}\log_{7}(9x - x^{2} - 13) > 0 \implies \lo

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$7$

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(D) For the function $f(x) = \log_{3}\log_{5}\log_{7}(9x - x^{2} - 13)$ to be defined, we require: $\log_{5}\log_{7}(9x - x^{2} - 13) > 0 \implies \log_{7}(9x - x^{2} - 13) > 1 \implies 9x - x^{2} - 13 > 7$ $x^{2} - 9x + 20 Thus, $m = 4$ and $n = 5$. For the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$, the eccentricity $e = \frac{n}{3} = \frac{5}{3}$. Since $e^{2} = 1 + \frac{b^{2}}{a^{2}}$, we have $\frac{25}{9} = 1 + \frac{b^{2}}{a^{2}} \implies \frac{b^{2}}{a^{2}} = \frac{16}{9} \implies \frac{b}{a} = \frac{4}{3} \implies b = \frac{4a}{3}$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{8m}{3} = \frac{32}{3}$. Substituting $b = \frac{4a}{3}$, we get $\frac{2(16a^{2}/9)}{a} = \frac{32}{3} \implies \frac{32a}{9} = \frac{32}{3} \implies a = 3$. Then $b = \frac{4(3)}{3} = 4$. Therefore, $b^{2} - a^{2} = 16 - 9 = 7$.

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 $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ have eccentricity $\frac{n}{3}$ and the length of the latus rectum $\frac{8m}{3}$. Then $b^{2} - a^{2}$ is equal to:

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