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### التمرين الأول (6 نقاط):

#### 1. **الحساب:**
Given: \(\log 40 + \log 25\), we use the property of logarithms:

\[
\log 40 + \log 25 = \log (40 \times 25) = \log 1000 = 3
\]

Thus, the result is \(3\).

#### 2. **حل المعادلة:**
Solve \(\ln(2x) = \ln 10\). Using the fact that \(\ln a = \ln b \Rightarrow a = b\):

\[
2x = 10 \quad \Rightarrow \quad x = \frac{10}{2} = 5
\]

#### 3. **حل المعادلة:**
Given \(e^{3x} = e^2\). Since the exponential function is one-to-one, we can equate the exponents:

\[
3x = 2 \quad \Rightarrow \quad x = \frac{2}{3}
\]

#### 4. **النهاية:**
\(\lim_{x \to 0} \frac{e^x - 1}{x}\). Using L'Hôpital's rule:

\[
\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = 1
\]

### التمرين الثاني (6 نقاط):

Given the sequence \(u_n = 4n - 3\), we proceed as follows:

#### 1. **تحقق أن:**
To prove \(u_n = 4n - 3\), it's already given. The formula defines the sequence explicitly.

#### 2. **المتتالية:**
The first term \(u_0\), second term \(u_1\), and sum of the first 18 terms:

\[
u_0 = 4(0) - 3 = -3, \quad u_1 = 4(1) - 3 = 1
\]

To find the sum \(S = u_0 + u_1 + \dots + u_{17}\), we use the sum formula for an arithmetic sequence:

\[
S_n = \frac{n}{2} \times (a_1 + a_n)
\]

Where \(a_1 = u_0 = -3\), \(a_n = u_{17} = 4(17) - 3 = 65\), and \(n = 18\):

\[
S_{18} = \frac{18}{2} \times (-3 + 65) = 9 \times 62 = 558
\]

### التمرين الثالث (6 نقاط):

The function \(f(x) = 2 - e^{-x}\) is defined. Now:

#### 1. **تحقق أن:**
Check \(f(\ln 2) = 0\):

\[
f(\ln 2) = 2 - e^{-\ln 2} = 2 - \frac{1}{2} = \frac{3}{2} \quad \text{(not zero, there seems to be a mistake)}
\]

Similarly for the limits:

\[
\lim_{x \to +\infty} f(x) = 2, \quad \lim_{x \to -\infty} f(x) = -\infty
\]

#### 2. **اشتقاق:**
Differentiate \(f(x)\):

\[
f'(x) = \frac{d}{dx} \left(2 - e^{-x}\right) = e^{-x}
\]

### التمرين الرابع (6 نقاط):

This problem involves probability with colored balls (3 white, 5 blue). 

#### 1. **احسب عدد التوزيعات الممكنة:**
Total number of ways to distribute 6 balls (3 white, 5 blue) in a sequence without distinction:

\[
\binom{6}{3} = \frac{6!}{3!3!} = 20
\]

#### 2. **احتمال الحصول:**
Probability of obtaining three blue balls when choosing 3 out of 5:

\[
P = \frac{\binom{3}{3}}{\binom{6}{3}} = \frac{1}{20}
\]

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Would you like to dive into the details of any of these steps? Here are some follow-up questions you might want to explore:

1. Can we apply another method for solving logarithmic equations?
2. How does L'Hôpital's rule work in more complex limits?
3. What are other properties of arithmetic sequences that can be useful?
4. Can you derive the differentiation of composite functions like \(f(x)\)?
5. How is probability affected if we add more balls of different colors?

### Tip:
In probability, breaking down problems into distinct cases often helps clarify the solution path!

This exam paper covers logarithmic calculations, solving exponential equations, evaluating limits, arithmetic sequences, differentiation, and probability. It provides step-by-step solutions for each problem, suitable for students in grades 10-12.

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