We are given the equation:

$\ln(7x) + 7 = \ln(3x + 2)$

Step 1: Isolate the logarithmic expressions

Start by subtracting 7 from both sides to isolate the logarithms:

$\ln(7x) = \ln(3x + 2) - 7$

Step 2: Eliminate the logarithms

For simpler equations, we can apply the logarithmic property: $\ln(a) = \ln(b) \implies a = b$ So we set the arguments of the logarithms equal to each other:

$7x = 3x + 2$

Step 3: Solve for $x$

Now solve the equation:

$7x - 3x = 2$

$4x = 2$

$x = \frac{2}{4} = \frac{1}{2}$

Step 4: Check for extraneous solutions

Make sure that the solution satisfies the original equation, particularly that the arguments inside the logarithms are positive. For $x = \frac{1}{2}$:

$7x = 7 \times \frac{1}{2} = \frac{7}{2} > 0$

$3x + 2 = 3 \times \frac{1}{2} + 2 = \frac{3}{2} + 2 = \frac{7}{2} > 0$

Both logarithmic arguments are positive, so $x = \frac{1}{2}$ is valid.

Thus, the solution is:

$x = \frac{1}{2}$

Would you like further details, or have any questions?

Related Questions:

1. How do you solve logarithmic equations with multiple terms?
2. What are the properties of logarithms used in solving equations?
3. Can logarithmic equations have no solutions?
4. How do you solve logarithmic inequalities?
5. What happens when the argument of a logarithm is negative or zero?

Tip:

Always check for extraneous solutions when solving logarithmic equations, as the domain of logarithms requires positive arguments.