If \( \ln (7 x+5)=2 \), then \( x=\square \)

To solve the equation \( \ln (7x + 5) = 2 \), we first exponentiate both sides to eliminate the natural logarithm. This gives us: \[ 7x + 5 = e^2 \] Next, we subtract 5 from both sides: \[ 7x = e^2 - 5 \] Now, we divide by 7 to isolate \( x \): \[ x = \frac{e^2 - 5}{7} \] Using an approximate value for \( e^2 \) (approximately 7.389), we can further compute: \[ x \approx \frac{7.389 - 5}{7} \approx \frac{2.389}{7} \approx 0.3413 \] Thus, the solution is \( x \approx 0.3413 \).