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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.3.1a

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

Verified step by step guidance
1
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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.
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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.
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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.
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Solving Exponential Equations Using Logs