To solve for $a$ and $b$ in the equation:

$\ln(5x + 10) = \ln(x - a) + b$

we’ll follow these steps:

Step 1: Remove the Natural Logarithm

Since we have a natural logarithm ($\ln$) on both sides, we can simplify by exponentiating both sides:

$5x + 10 = e^{\ln(x - a) + b}$

This simplifies further to:

$5x + 10 = e^{\ln(x - a)} \cdot e^b$

Since $e^{\ln(x - a)} = x - a$, the equation becomes:

$5x + 10 = (x - a)e^b$

Step 2: Solve for $a$ and $b$

To determine $a$ and $b$, let's equate coefficients for $x$ and constant terms separately:

1. For the coefficient of $x$: $5 = e^b$ Taking the natural logarithm of both sides gives: $b = \ln(5)$

2. For the constant term: $10 = -a \cdot e^b$ Substitute $e^b = 5$ from the previous step: $10 = -a \cdot 5$ Solving for $a$: $a = -2$

Solution

The values for $a$ and $b$ are:

$a = -2 \quad \text{and} \quad b = \ln(5)$

Would you like further details on the steps or have any questions?

Here are some related questions that might deepen your understanding:

1. How would the solution change if the original equation included a different constant?
2. What happens if we have an additional logarithmic term on the right side?
3. Can we apply similar methods to equations with other bases of logarithms?
4. How does exponentiating both sides help when solving logarithmic equations?
5. What are the implications if the argument inside the logarithm is negative?

Tip: When solving logarithmic equations, ensure that arguments inside logarithms remain positive for all real values of the variables involved.