Let's solve each equation step-by-step.

Part (a):

The equation is: $\log_3(7x + 1) - \log_3(x - 5) = 2$

Using the properties of logarithms, specifically the subtraction rule $\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)$, we can rewrite this as: $\log_3\left(\frac{7x + 1}{x - 5}\right) = 2$

Now, rewrite this logarithmic equation in exponential form: $\frac{7x + 1}{x - 5} = 3^2$ $\frac{7x + 1}{x - 5} = 9$

Now, cross-multiply to solve for $x$: $7x + 1 = 9(x - 5)$ $7x + 1 = 9x - 45$ $1 + 45 = 9x - 7x$ $46 = 2x$ $x = 23$

So, the solution for part (a) is: $x = 23$

Part (b):

The equation is: $\log_5(x - 24) + \log_5(x) = 2$

Using the addition rule for logarithms $\log_b(A) + \log_b(B) = \log_b(AB)$, we rewrite it as: $\log_5((x - 24) \cdot x) = 2$ $\log_5(x^2 - 24x) = 2$

Rewrite this equation in exponential form: $x^2 - 24x = 5^2$ $x^2 - 24x = 25$

Now, bring all terms to one side of the equation: $x^2 - 24x - 25 = 0$

This is a quadratic equation. Solve it using the quadratic formula: $x = \frac{-(-24) \pm \sqrt{(-24)^2 - 4 \cdot 1 \cdot (-25)}}{2 \cdot 1}$ $x = \frac{24 \pm \sqrt{576 + 100}}{2}$ $x = \frac{24 \pm \sqrt{676}}{2}$ $x = \frac{24 \pm 26}{2}$

This gives two solutions:

1. $x = \frac{24 + 26}{2} = 25$
2. $x = \frac{24 - 26}{2} = -1$

However, $x = -1$ is not a valid solution because we cannot take the logarithm of a negative number. Therefore, the only solution is: $x = 25$

Summary of Solutions:

• For (a), $x = 23$
• For (b), $x = 25$

Would you like further details or explanations on any steps?

Related Questions:

1. What is the general rule for converting logarithmic equations to exponential form?
2. How can we verify if a solution to a logarithmic equation is valid?
3. What are the restrictions on values for variables in logarithmic functions?
4. How does the quadratic formula work, and why does it give two possible solutions?
5. What other properties of logarithms are commonly used in solving equations?

Tip:

Always check if potential solutions satisfy the domain requirements of the logarithmic functions, as logarithms are only defined for positive arguments.