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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
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7장, 문제 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 \(t\).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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Exponential functions involve variables in the exponent, such as e^t. Understanding their properties, like how e^(a+b) = e^a * e^b, helps simplify and solve equations involving exponentials.
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The substitution method replaces a complex expression with a simpler variable to transform the equation into a more familiar form, such as turning e^t into x, making it easier to solve.
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After substitution, the equation often becomes quadratic. Knowing how to solve quadratic equations using factoring, the quadratic formula, or completing the square is essential to find the variable's values.
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Solving Logarithmic Equations