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