Find the solution to the equation, \( 2 e^{3 y+8}-11 e^{5-10 y}=0 \) (correct to 4 decimal places) \( \} \) y \( =\square \)

To solve the equation \( 2 e^{3 y+8} - 11 e^{5-10 y} = 0 \), we can start by rewriting it in a more manageable form. First, isolate one of the exponential terms: \[ 2 e^{3 y + 8} = 11 e^{5 - 10y} \] Next, divide both sides by \( e^{5 - 10y} \): \[ \frac{2 e^{3y + 8}}{e^{5 - 10y}} = 11 \] This simplifies to: \[ 2 e^{3y + 8 - (5 - 10y)} = 11 \] which further simplifies to: \[ 2 e^{13y + 3} = 11 \] Now, divide both sides by 2: \[ e^{13y + 3} = \frac{11}{2} \] Take the natural logarithm of both sides: \[ 13y + 3 = \ln\left(\frac{11}{2}\right) \] Now, isolate \( y \): \[ 13y = \ln\left(\frac{11}{2}\right) - 3 \] \[ y = \frac{\ln\left(\frac{11}{2}\right) - 3}{13} \] Calculating \( \ln\left(\frac{11}{2}\right) \): \[ \ln\left(\frac{11}{2}\right) \approx 2.3979 \] Substituting back: \[ y \approx \frac{2.3979 - 3}{13} \approx \frac{-0.6021}{13} \approx -0.0463 \] Thus, the solution to the equation, correct to four decimal places, is: \[ y \approx -0.0463 \]