Find the solution to the equation, \( 2 e^{3 y+8}-11 e^{5-10 y}=0 \) (correct to 4 decimal places) \( \} \mathrm{y}= \)
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To solve the equation \( 2 e^{3y+8} - 11 e^{5-10y} = 0 \), we can start by rearranging it: \[ 2 e^{3y+8} = 11 e^{5-10y}. \] Dividing both sides by \( e^{5-10y} \): \[ \frac{2 e^{3y+8}}{e^{5-10y}} = 11. \] This simplifies to: \[ 2 e^{3y + 8 + 10y - 5} = 11, \] \[ 2 e^{13y + 3} = 11. \] Now, isolate the exponential term: \[ e^{13y + 3} = \frac{11}{2}. \] Taking the natural logarithm of both sides gives: \[ 13y + 3 = \ln\left(\frac{11}{2}\right). \] Now, solve for \( y \): \[ 13y = \ln\left(\frac{11}{2}\right) - 3, \] \[ y = \frac{\ln\left(\frac{11}{2}\right) - 3}{13}. \] Calculating this yields: \[ y \approx \frac{\ln(5.5) - 3}{13} \approx \frac{1.7047 - 3}{13} \approx \frac{-1.2953}{13} \approx -0.0996. \] Thus, rounded to four decimal places, the solution is: \( y \approx -0.0996. \)