Question

Which of the following correctly solves the equation log3(2x−1)=log3(3x)
?(1 point)
Responses

x=1
x equals 1

x=−1
x equals negative 1

x=1/5
x equals Start Fraction 1 over 5 End Fraction

no solution
no solution

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 \( a = c \) (as long as \( a > 0 \) and \( c > 0 \)).

Setting the arguments equal to each other, we get:

\[
2x - 1 = 3x
\]

Next, we will rearrange the equation:

\[
2x - 3x = 1 \\
-x = 1 \\
x = -1
\]

Now we need to check if \( x = -1 \) is a valid solution by substituting it back into the original logarithmic functions:

1. For \( 2x - 1 \):
\[
2(-1) - 1 = -2 - 1 = -3
\]
Since logarithms of negative numbers are undefined, \( \log_3(2(-1) - 1) \) is undefined.

2. For \( 3x \):
\[
3(-1) = -3
\]
Again, \( \log_3(-3) \) is undefined.

Since both sides are undefined for \( x = -1 \), it is not a valid solution.

Next, we need to check if there are any other possible values for \( x \). The arguments inside the logarithms must be greater than zero:

\[
2x - 1 > 0 \quad (1)
\]
\[
3x > 0 \quad (2)
\]

From inequality (1):
\[
2x > 1 \\
x > \frac{1}{2}
\]

From inequality (2):
\[
3x > 0 \\
x > 0
\]

Both conditions imply \( x > \frac{1}{2} \). Since \( x = -1 \) does not satisfy this condition and we haven't found any other solutions, we conclude that the original equation has **no solution** when considering the requirements for logarithms.

Thus, the correct response is:

**no solution**.