Solve the following equations:
(a) e^(2x) − 3e^x + 2 = 0
(b) log(base2) 3 + log(base2) x = log(base2) 5 + log(base2)(x − 2)

Let's solve each equation step by step:

### (a) \( e^{2x} - 3e^x + 2 = 0 \)

#### Step 1: Substitute \( y = e^x \)
This substitution simplifies the equation. Since \( e^{2x} = (e^x)^2 \), we rewrite the equation as:
\[
y^2 - 3y + 2 = 0
\]

#### Step 2: Solve the quadratic equation
Now we solve the quadratic equation \( y^2 - 3y + 2 = 0 \). Factoring gives:
\[
(y - 1)(y - 2) = 0
\]
So, \( y = 1 \) or \( y = 2 \).

#### Step 3: Substitute back \( y = e^x \)
Since \( y = e^x \), we have two cases:
- \( e^x = 1 \), which gives \( x = \ln(1) = 0 \).
- \( e^x = 2 \), which gives \( x = \ln(2) \).

Thus, the solutions to the first equation are:
\[
x = 0 \quad \text{or} \quad x = \ln(2)
\]

---

### (b) \( \log_2(3) + \log_2(x) = \log_2(5) + \log_2(x - 2) \)

#### Step 1: Use the logarithmic property
Recall that \( \log_b(A) + \log_b(B) = \log_b(A \cdot B) \). Apply this property to both sides of the equation:
\[
\log_2(3x) = \log_2(5(x - 2))
\]

#### Step 2: Remove the logarithms
Since the logarithms have the same base, we can equate the arguments:
\[
3x = 5(x - 2)
\]

#### Step 3: Solve the linear equation
Expand the right-hand side:
\[
3x = 5x - 10
\]
Rearrange the terms:
\[
3x - 5x = -10 \implies -2x = -10
\]
Solving for \( x \):
\[
x = 5
\]

#### Step 4: Check for extraneous solutions
We need to check that \( x - 2 > 0 \), which gives \( x > 2 \). Since \( x = 5 \), it satisfies this condition.

Thus, the solution to the second equation is:
\[
x = 5
\]

---

### Final Solutions:
(a) \( x = 0 \) or \( x = \ln(2) \)

(b) \( x = 5 \)

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Would you like any further details or explanations on any step?

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Here are 5 related questions for further practice:
1. Solve \( e^{3x} - 4e^x + 3 = 0 \).
2. Solve \( \log_3(x^2) = \log_3(5x) \).
3. Solve \( e^{2x} - 5e^x + 6 = 0 \).
4. Solve \( \log_5(x+1) + \log_5(x-1) = 1 \).
5. Solve \( e^x + e^{-x} = 5 \).

**Tip**: When solving logarithmic equations, always check for domain restrictions like \( x > 0 \).

Solve the following equations: (a) e^(2x) − 3e^x + 2 = 0 (b) log(base2) 3 + log(base2) x = log(base2) 5 + log(base2)(x − 2)

Learn how to solve the exponential equation e^(2x) - 3e^x + 2 = 0 and the logarithmic equation log(base2) 3 + log(base2) x = log(base2) 5 + log(base2)(x - 2). This problem covers key algebraic concepts such as solving quadratic equations through substitution and using logarithmic properties. Ideal for students in Grades 9-11.

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