If the area of the region bounded by 16x^2 - 9y^2 = 144 and 8x - 3y...

The given equations are:

Hyperbola: $16x^2 - 9y^2 = 144 \Rightarrow \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

Line: $8x - 3y = 24 \Rightarrow y = \dfrac{8}{3}(x - 3)$

To find the points of intersection, substitute $y$ from the line equation into the hyperbola equation:

$16x^2 - 9\left(\dfrac{8}{3}(x - 3)\right)^2 = 144$

$16x^2 - 64(x^2 - 6x + 9) = 144$

$x^2 - 4(x^2 - 6x + 9) = 9$

$-3x^2 + 24x - 45 = 0$

$x^2 - 8x + 15 = 0$

$(x - 3)(x - 5) = 0 \Rightarrow x = 3, x = 5$

For $x \in [3, 5]$, the upper curve is the hyperbola $y = \dfrac{4}{3}\sqrt{x^2 - 9}$ and the lower curve is the line $y = \dfrac{8}{3}(x - 3)$.

The area $A$ of the bounded region is:

$A = \int_{3}^{5} \left( \dfrac{4}{3}\sqrt{x^2 - 9} - \dfrac{8}{3}(x - 3) \right) dx$

Using the standard integral formula $\int \sqrt{x^2 - a^2} dx = \dfrac{x}{2}\sqrt{x^2 - a^2} - \dfrac{a^2}{2}\ln|x + \sqrt{x^2 - a^2}|$:

$\int_{3}^{5} \sqrt{x^2 - 9} dx = \left[ \dfrac{x}{2}\sqrt{x^2 - 9} - \dfrac{9}{2}\ln|x + \sqrt{x^2 - 9}| \right]_{3}^{5}$

$= \left( \dfrac{5}{2}(4) - \dfrac{9}{2}\ln 9 \right) - \left( 0 - \dfrac{9}{2}\ln 3 \right)$

$= 10 - 9\ln 3 + \dfrac{9}{2}\ln 3 = 10 - \dfrac{9}{2}\ln 3$

For the linear part:

$\int_{3}^{5} (x - 3) dx = \left[ \dfrac{(x - 3)^2}{2} \right]_{3}^{5} = \dfrac{4}{2} - 0 = 2$

Substituting these back into the area equation:

$A = \dfrac{4}{3}\left( 10 - \dfrac{9}{2}\ln 3 \right) - \dfrac{8}{3}(2)$

$A = \dfrac{40}{3} - 6\ln 3 - \dfrac{16}{3}$

$A = 8 - 6\ln 3$

We need to find the value of $3(A + 6\log_e 3)$:

$A + 6\ln 3 = 8$

$3(A + 6\ln 3) = 3 \times 8 = 24$

Answer: $24$