405e^(-3t)-540e^(-2t)+135e^(-t)=0 Solution Solve for t over the real numbers: 40

Question

405e^(-3t)-540e^(-2t)+135e^(-t)=0

Explanation / Answer

Solve for t over the real numbers: 405/e^(3 t)-540/e^(2 t)+135/e^t = 0 Factor the left hand side. The left hand side factors into a product with four terms: (135 (e^t-3) (e^t-1))/e^(3 t) = 0 Divide both sides by a constant to simplify the equation. Divide both sides by 135: ((e^t-3) (e^t-1))/e^(3 t) = 0 Solve each term in the product separately. Split into three equations: e^(-3 t) = 0 or e^t-3 = 0 or e^t-1 = 0 Look at the first equation: Show the equation has no solution. e^(-3 t) = 0 has no solution since for all t element R, e^(-3 t)>0 and : e^t-3 = 0 or e^t-1 = 0 Isolate terms with t to the left hand side. Add 3 to both sides: e^t = 3 or e^t-1 = 0 Eliminate the exponential from the left hand side. Take the natural logarithm of both sides: t = log(3) or e^t-1 = 0 Look at the second equation: Isolate terms with t to the left hand side. Add 1 to both sides: t = log(3) or e^t = 1 Eliminate the exponential from the left hand side. Take the natural logarithm of both sides: Answer: | | t = log(3) or t = 0

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