$$ \sqrt{a-2} $$=2$$ \sqrt{b+1} $$   equation 1                                                                                                                                                      $$ 3^{\left(c-4\right)} $$+1=$$ a\cdot b $$          equation 2                                                                                                                                                       $$ \log_{b}\left(d\right) $$$$ -1=\log_{b}\left(d-16\right) $$  equation 3                                                                                                                               $$ \log\left(a-4\right) $$=1                     equation 4                                                                                                                                          $$ \int_{aa}^{d}\!b\,\mathrm{dt-} $$m.c.m($$ a\cdot b\cdot c\cdot d) $$=? what will be the answer ? is there any possibility that answer would be from this options (32,16,2,240,224,14)

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.

Learn how to solve complex equations involving square roots, exponential functions, logarithms, and definite integrals step-by-step. This problem covers key mathematical concepts and formulas, including properties of square roots and logarithms, and how to evaluate definite integrals. Suitable for high school students studying advanced algebra and calculus concepts.

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