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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.3.5

5. e^(2t)-3e^t = 0

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Rewrite the equation \(e^{2t} - 3e^t = 0\) by recognizing that \(e^{2t}\) can be expressed as \((e^t)^2\). This allows us to treat the equation like a quadratic in terms of \(e^t\).
Let \(x = e^t\). Substitute this into the equation to get \(x^2 - 3x = 0\).
Factor the quadratic equation: \(x(x - 3) = 0\). This gives two possible solutions for \(x\): \(x = 0\) or \(x = 3\).
Recall that \(x = e^t\), and since \(e^t\) is never zero for any real \(t\), discard \(x = 0\). Focus on \(e^t = 3\).
Solve for \(t\) by taking the natural logarithm of both sides: \(t = \ln(3)\). This gives the solution for \(t\).

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