Solve: \( \log _{7}(2 x-1)=2 \) A \( 64 \frac{1}{2} \) B \( 63 \frac{1}{2} \) C 25 (D) 24

To solve the equation \( \log _{7}(2 x-1)=2 \), we can rewrite it in exponential form. This means we have \( 2x - 1 = 7^2 \). Calculating \( 7^2 \) gives us 49. So the equation becomes: \[ 2x - 1 = 49 \] Now, add 1 to both sides: \[ 2x = 50 \] Next, divide by 2: \[ x = 25 \] So the correct answer is C 25. Now, for some additional insights... Knowing the properties of logarithms can make solving such problems a breeze! For instance, remember that \( \log_b(a) = c \) translates to \( b^c = a \). This is a handy tool in your mathematical belt! Also, practicing with different bases and equations can sharpen your skills! You can try changing the base to find equivalent logarithmic values, or even experiment with more complex equations involving logarithm rules. It’s like a math puzzle!