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: $x = 0 \quad \text{or} \quad x = \ln(2)$

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(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: $x = 5$

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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 $x > 0$.