Let's analyze and solve the system of equations step by step to determine the final answer.

Equation 1:

$\sqrt{a - 2} = 2\sqrt{b + 1}$ Square both sides: $a - 2 = 4(b + 1)$ $a - 2 = 4b + 4$ $a = 4b + 6 \quad \text{(Equation 1.1)}$

Equation 2:

$3^{c-4} + 1 = a \cdot b$ Subtract 1 from both sides: $3^{c-4} = a \cdot b - 1 \quad \text{(Equation 2.1)}$

Equation 3:

$\log_b(d) - 1 = \log_b(d-16)$ Add 1 to both sides: $\log_b(d) = \log_b(d-16) + 1$ Convert this logarithmic equation into its exponential form: $d = b \cdot (d - 16)$ Expanding and rearranging: $d = bd - 16b$ $d(1 - b) = -16b$ $d = \frac{16b}{b-1} \quad \text{(Equation 3.1)}$

Equation 4:

$\log(a - 4) = 1$ This can be rewritten as: $a - 4 = 10^1$ $a - 4 = 10$ $a = 14 \quad \text{(Equation 4.1)}$

Substitute $a = 14$ into Equation 1.1:

$14 = 4b + 6$ $4b = 8$ $b = 2$

Substitute $a = 14$ and $b = 2$ into Equation 2.1:

$3^{c-4} = 14 \cdot 2 - 1$ $3^{c-4} = 28 - 1$ $3^{c-4} = 27$ Since $27 = 3^3$, we get: $c - 4 = 3$ $c = 7$

Substitute $b = 2$ into Equation 3.1:

$d = \frac{16 \cdot 2}{2 - 1}$ $d = 32$

Final Expression to Evaluate:

$\int_{a}^{d} b \, dt - \text{m.c.m}(a \cdot b \cdot c \cdot d)$ Since $b$ is a constant: $\int_{14}^{32} 2 \, dt = 2(t) \Big|_{14}^{32} = 2(32 - 14) = 2 \times 18 = 36$

Now, let's find the m.c.m (least common multiple, LCM) of $a \cdot b \cdot c \cdot d$: $a = 14, \, b = 2, \, c = 7, \, d = 32$ $a \cdot b \cdot c \cdot d = 14 \times 2 \times 7 \times 32 = 6272$ The LCM of 6272 is 6272 itself.

Final Result:

$36 - 6272 = -6236$

The possible answers provided (32, 16, 2, 240, 224, 14) do not match the result of $-6236$. Thus, the correct answer is not among the given options.

Would you like more details or have any further questions?

Related Questions:

1. How can logarithmic properties simplify solving equations?
2. What is the relationship between logarithms and exponential functions?
3. How do you find the LCM of a set of numbers?
4. What are the key properties of definite integrals?
5. How does squaring both sides of an equation affect its solutions?

Tip: Always check the domains of logarithmic and square root functions before solving the equations.