Question

Solve log3(2x-1)=log3(3x)

Answer 1

To solve the equation \( \log_3(2x - 1) = \log_3(3x) \), we can use the property of logarithms that states if \( \log_b(a) = \log_b(c) \), then \( a = c \), provided that \( a \) and \( c \) are within the domain of the logarithm function.

Applying that property, we set the arguments equal to each other:

\[
2x - 1 = 3x
\]

Now, we can solve for \( x \). Rearranging the equation gives:

\[
2x - 3x = 1
\]

This simplifies to:

\[
-x = 1
\]

Now, multiplying both sides by -1:

\[
x = -1
\]

Next, we need to verify if this solution is valid by checking if it falls within the domain of the logarithmic functions. The arguments of the logarithms must be greater than zero:

1. For \( 2x - 1 \):
\[
2(-1) - 1 = -2 - 1 = -3 \quad \text{(not valid)}
\]

2. For \( 3x \):
\[
3(-1) = -3 \quad \text{(not valid)}
\]

Since both arguments \( (2x - 1) \) and \( (3x) \) are negative when \( x = -1 \), this solution is not valid in the context of the logarithmic functions.

Thus, there are **no valid solutions** to the equation \( \log_3(2x - 1) = \log_3(3x) \).