$$ \sqrt{a-2} $$=2$$ \sqrt{b+1} $$   equation 1                                                                                                                                                      $$ 3^{\left(c-4\right)} $$+1=$$ ab $$          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($$ abcd) $$=? what will be the answer ? is there any possibility that answer would be from this options (32,16,2,240,224,14)

Let's solve the system of equations step by step to find the correct answer.

### Step 1: Solve Equation 4

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

### Step 2: Substitute $a = 14$ into Equation 1

Given:
\[
\sqrt{a - 2} = 2\sqrt{b + 1}
\]
Substitute $a = 14$:
\[
\sqrt{14 - 2} = 2\sqrt{b + 1}
\]
\[
\sqrt{12} = 2\sqrt{b + 1}
\]
Square both sides:
\[
12 = 4(b + 1)
\]
\[
12 = 4b + 4
\]
\[
4b = 8
\]
\[
b = 2 \quad \text{(Equation 1.1)}
\]

### Step 3: Substitute $a = 14$ and $b = 2$ into Equation 2

Given:
\[
3^{c-4} + 1 = ab
\]
Substitute $a = 14$ and $b = 2$:
\[
3^{c-4} + 1 = 14 \times 2
\]
\[
3^{c-4} + 1 = 28
\]
Subtract 1 from both sides:
\[
3^{c-4} = 27
\]
Since $27 = 3^3$, we get:
\[
c - 4 = 3
\]
\[
c = 7 \quad \text{(Equation 2.1)}
\]

### Step 4: Substitute $b = 2$ into Equation 3

Given:
\[
\log_b(d) - 1 = \log_b(d - 16)
\]
Add 1 to both sides:
\[
\log_b(d) = \log_b(d - 16) + 1
\]
Convert this logarithmic equation into its exponential form:
\[
d = b \cdot (d - 16)
\]
Substitute $b = 2$:
\[
d = 2(d - 16)
\]
\[
d = 2d - 32
\]
\[
d - 2d = -32
\]
\[
-d = -32
\]
\[
d = 32 \quad \text{(Equation 3.1)}
\]

### Step 5: Evaluate the Given Expression

We need to compute:
\[
\int_{a}^{d} b \, dt - \text{m.c.m}(ab \cdot c \cdot d)
\]

#### Compute the Integral
\[
\int_{14}^{32} 2 \, dt = 2 \times (32 - 14) = 2 \times 18 = 36
\]

#### Compute the m.c.m (LCM) of $ab \cdot c \cdot d$
The product $ab \cdot c \cdot d$ is:
\[
ab \cdot c \cdot d = 14 \times 2 \times 7 \times 32 = 6272
\]
The least common multiple (LCM) of $6272$ is $6272$ itself.

### Final Calculation

\[
\int_{a}^{d} b \, dt - \text{m.c.m}(ab \cdot c \cdot d) = 36 - 6272 = -6236
\]

### Conclusion
The final answer, $-6236$, does not match any of the provided options (32, 16, 2, 240, 224, 14). Therefore, the correct answer is not among the options given.

Would you like more details or have any further questions?

**Related Questions:**
1. How do you determine the LCM of a set of numbers?
2. What is the significance of solving logarithmic equations?
3. Why is squaring both sides of an equation sometimes necessary?
4. How does the definite integral apply to calculating areas?
5. What are some common mistakes when solving systems of equations?

**Tip:** Always verify each step carefully to ensure there are no calculation errors, especially when working with logarithms and exponents.

Explore the step-by-step solution to equations involving logarithms, exponential equations, and definite integrals. Learn how to apply logarithm properties and solve exponential equations effectively. Suitable for advanced high school students.

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