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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.3.1a

In Exercises 1–4, solve for t.
1. a. e^(-0.3t) = 27

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Identify the equation given: \(e^{-0.3t} = 27\).
To solve for \(t\), take the natural logarithm (ln) of both sides to undo the exponential function: \(\ln\left(e^{-0.3t}\right) = \ln(27)\).
Use the logarithm property \(\ln\left(e^x\right) = x\) to simplify the left side: \(-0.3t = \ln(27)\).
Isolate \(t\) by dividing both sides of the equation by \(-0.3\): \(t = \frac{\ln(27)}{-0.3}\).
At this point, you can evaluate \(\ln(27)\) using a calculator and then perform the division to find the value of \(t\).

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주요 개념

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

Exponential Functions

An exponential function has the form f(t) = a^t, where the variable is in the exponent. Understanding how to manipulate and interpret these functions is essential for solving equations like e^(-0.3t) = 27, where the unknown appears as an exponent.
추천 영상:
6:13
Exponential Functions

Natural Logarithm (ln)

The natural logarithm is the inverse of the exponential function with base e. Applying the natural logarithm to both sides of an equation like e^(-0.3t) = 27 allows us to 'bring down' the exponent and solve for t.
추천 영상:
05:18
Derivative of the Natural Logarithmic Function

Solving Exponential Equations

Solving exponential equations involves isolating the exponential term and then using logarithms to solve for the variable in the exponent. This process often requires properties of logarithms and careful algebraic manipulation.
추천 영상:
5:47
Solving Exponential Equations Using Logs