Let’s solve the logarithmic equations step by step:

(a) $\log_{10}(x) + \log_{10}(x+9) = 1$

Step 1: Combine the logarithmic terms Using the property of logarithms $\log_b(A) + \log_b(B) = \log_b(A \cdot B)$, we can combine the terms: $\log_{10}(x) + \log_{10}(x+9) = \log_{10}(x(x+9))$ Thus, the equation becomes: $\log_{10}(x(x+9)) = 1$

Step 2: Rewrite in exponential form Rewriting the logarithmic equation $\log_{10}(x(x+9)) = 1$ in its exponential form: $x(x+9) = 10^1$ $x(x+9) = 10$

Step 3: Expand and simplify Expand the left-hand side: $x^2 + 9x - 10 = 0$

Step 4: Solve the quadratic equation Solve $x^2 + 9x - 10 = 0$ using factoring or the quadratic formula. Factoring: $(x + 10)(x - 1) = 0$ $x = -10 \quad \text{or} \quad x = 1$

Step 5: Check for valid solutions Logarithms are only defined for positive arguments, so $x > 0$. Therefore, $x = -10$ is invalid.

Solution for (a): $x = 1$

(b) $\log_2(x-1) - \log_2(x-5) = 3$

Step 1: Combine the logarithmic terms Using the property of logarithms $\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)$, we can combine the terms: $\log_2(x-1) - \log_2(x-5) = \log_2\left(\frac{x-1}{x-5}\right)$ Thus, the equation becomes: $\log_2\left(\frac{x-1}{x-5}\right) = 3$

Step 2: Rewrite in exponential form Rewriting the logarithmic equation $\log_2\left(\frac{x-1}{x-5}\right) = 3$ in its exponential form: $\frac{x-1}{x-5} = 2^3$ $\frac{x-1}{x-5} = 8$

Step 3: Solve for $x$ Multiply through by $(x-5)$ to eliminate the fraction: $x - 1 = 8(x - 5)$ Expand the right-hand side: $x - 1 = 8x - 40$ Simplify: $-1 + 40 = 8x - x$ $39 = 7x$ $x = \frac{39}{7}$

Step 4: Check for valid solutions For logarithms to be valid, both arguments $x-1$ and $x-5$ must be positive: $x - 1 > 0 \quad \text{and} \quad x - 5 > 0$ $x > 1 \quad \text{and} \quad x > 5$ Thus, $x > 5$. Since $\frac{39}{7} \approx 5.57 > 5$, it is valid.

Solution for (b): $x = \frac{39}{7}$

Final Answers:

(a) $x = 1$
(b) $x = \frac{39}{7}$

Do you have questions about these solutions or need further details? Here are some related questions for practice:

1. How would you solve a logarithmic equation with a quadratic argument, like $\log_2(x^2 + 5x + 6) = 2$?
2. What happens if a logarithmic equation has no valid solutions due to domain restrictions?
3. Can you derive the general properties of logarithms from their exponential form?
4. How would the solutions change if the logarithm bases were different, like $\ln(x) + \ln(x+1)$?
5. How do you graph $\log_2(x-1)$ to verify solutions?

Tip: Always check the domain of logarithmic equations after solving, as not all solutions may be valid!