Let's solve each equation step by step:

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a) $4^{2x+5} = 32^{4 - x}$

Step 1: Express both sides with the same base.

We know that: $4 = 2^2 \quad \text{and} \quad 32 = 2^5$ So, we can rewrite the equation as: $(2^2)^{2x+5} = (2^5)^{4 - x}$

Step 2: Simplify the exponents using the power rule $(a^m)^n = a^{m \cdot n}$.

$2^{2(2x + 5)} = 2^{5(4 - x)}$ Simplify the exponents: $2^{4x + 10} = 2^{20 - 5x}$

Step 3: Since the bases are the same, set the exponents equal to each other:

$4x + 10 = 20 - 5x$

Step 4: Solve for $x$.

Add $5x$ to both sides: $4x + 5x + 10 = 20$ $9x + 10 = 20$ Subtract 10 from both sides: $9x = 10$ Divide by 9: $x = \frac{10}{9} \approx 1.111$

Step 5: Check for extraneous solutions.

Since both sides are exponential functions with positive bases, there are no restrictions, so the solution is valid.

Thus, the solution is: $x \approx 1.111$

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b) $\log_3 2 + 2\log_3 x = \log_3 (7x - 3)$

Step 1: Apply the logarithmic properties.

We use the property $n \log_b a = \log_b a^n$ to simplify the second term on the left-hand side: $\log_3 2 + \log_3 x^2 = \log_3 (7x - 3)$

Step 2: Combine the logarithms on the left-hand side using the property $\log_b a + \log_b c = \log_b (ac)$:

$\log_3 (2x^2) = \log_3 (7x - 3)$

Step 3: Since the logarithms are equal, their arguments must be equal as well:

$2x^2 = 7x - 3$

Step 4: Rearrange the equation to form a quadratic equation:

$2x^2 - 7x + 3 = 0$

Step 5: Solve the quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 2$, $b = -7$, and $c = 3$. Substituting into the formula: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$ $x = \frac{7 \pm \sqrt{49 - 24}}{4}$ $x = \frac{7 \pm \sqrt{25}}{4}$ $x = \frac{7 \pm 5}{4}$

So we have two possible solutions: $x = \frac{7 + 5}{4} = \frac{12}{4} = 3$ or $x = \frac{7 - 5}{4} = \frac{2}{4} = 0.5$

Step 6: Check for extraneous solutions.

For $x = 0.5$, the argument $7x - 3 = 7(0.5) - 3 = 3.5 - 3 = 0.5$, which is positive, so this is valid.

For $x = 3$, the argument $7x - 3 = 7(3) - 3 = 21 - 3 = 18$, which is also positive, so this is valid.

Thus, both solutions are valid.

The solutions are: $x = 3 \quad \text{or} \quad x = 0.5$

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Final Answers:

a) $x \approx 1.111$

b) $x = 3$ or $x = 0.5$

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Would you like more details on any of the steps? Let me know if you have any questions.

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Related Questions:

1. How do you solve exponential equations with different bases?
2. What are the properties of logarithms and how are they used in solving equations?
3. How do you check for extraneous solutions in logarithmic and exponential equations?
4. How do you use the quadratic formula to solve for $x$?
5. What are the steps in simplifying logarithmic expressions?

Tip:

When solving logarithmic equations, always remember to check if the arguments of the logarithms are positive, as logarithms are only defined for positive real numbers.