If the area of the region bounded by 16x^2 - 9y^2 = 144 and 8x - 3y = 24 is A, then 3(A + 6 log_e(3)) is equal to .

The given equations are:

Hyperbola: $16x^2 - 9y^2 = 144 \Rightarrow \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$16x2−9y2=144⇒9x2​−16y2​=1

Line: $8x - 3y = 24 \Rightarrow y = \dfrac{8}{3}(x - 3)$8x−3y=24⇒y=38​(x−3)

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

$$16x^2 - 9\left(\dfrac{8}{3}(x - 3)\right)^2 = 144$$16x2−9(38​(x−3))2=144

$$16x^2 - 64(x^2 - 6x + 9) = 144$$16x2−64(x2−6x+9)=144

$$x^2 - 4(x^2 - 6x + 9) = 9$$x2−4(x2−6x+9)=9

$$-3x^2 + 24x - 45 = 0$$−3x2+24x−45=0

$$x^2 - 8x + 15 = 0$$x2−8x+15=0

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

For $x \in [3, 5]$x∈[3,5], the upper curve is the hyperbola $y = \dfrac{4}{3}\sqrt{x^2 - 9}$y=34​x2−9​ and the lower curve is the line $y = \dfrac{8}{3}(x - 3)$y=38​(x−3).

The area $A$A of the bounded region is:

$$A = \int_{3}^{5} \left( \dfrac{4}{3}\sqrt{x^2 - 9} - \dfrac{8}{3}(x - 3) \right) dx$$A=∫35​(34​x2−9​−38​(x−3))dx

Using the standard integral formula $\displaystyle \int \sqrt{x^2 - a^2} dx = \dfrac{x}{2}\sqrt{x^2 - a^2} - \dfrac{a^2}{2}\ln|x + \sqrt{x^2 - a^2}|$∫x2−a2​dx=2x​x2−a2​−2a2​ln∣x+x2−a2​∣:

$$\int_{3}^{5} \sqrt{x^2 - 9} dx = \left[ \dfrac{x}{2}\sqrt{x^2 - 9} - \dfrac{9}{2}\ln|x + \sqrt{x^2 - 9}| \right]_{3}^{5}$$∫35​x2−9​dx=[2x​x2−9​−29​ln∣x+x2−9​∣]35​

$$= \left( \dfrac{5}{2}(4) - \dfrac{9}{2}\ln 9 \right) - \left( 0 - \dfrac{9}{2}\ln 3 \right)$$=(25​(4)−29​ln9)−(0−29​ln3)

$$= 10 - 9\ln 3 + \dfrac{9}{2}\ln 3 = 10 - \dfrac{9}{2}\ln 3$$=10−9ln3+29​ln3=10−29​ln3

For the linear part:

$$\int_{3}^{5} (x - 3) dx = \left[ \dfrac{(x - 3)^2}{2} \right]_{3}^{5} = \dfrac{4}{2} - 0 = 2$$∫35​(x−3)dx=[2(x−3)2​]35​=24​−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=34​(10−29​ln3)−38​(2)

$$A = \dfrac{40}{3} - 6\ln 3 - \dfrac{16}{3}$$A=340​−6ln3−316​

$$A = 8 - 6\ln 3$$A=8−6ln3

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

$$A + 6\ln 3 = 8$$A+6ln3=8

$$3(A + 6\ln 3) = 3 \times 8 = 24$$3(A+6ln3)=3×8=24

Answer: $24$24