Problem 1 - Thomas Calculus

Question

solve for \(t.\) \begin{equation}\quad \text { a. }e^{-0.3 t}=27 \quad \text { b. } e^{k t}=\frac{1}{2} \quad \text { c. } e^{(\ln 0.2) t}=0.4\end{equation}

Step-by-Step Solution

Answer

a. \( t = \frac{\ln(27)}{-0.3} \); b. \( t = \frac{\ln(1/2)}{k} \); c. \( t = \frac{\ln(0.4)}{\ln 0.2} \).

Step 1: Solve Part (a)

For the equation \( e^{-0.3t} = 27 \), take the natural logarithm of both sides to remove the exponential function. This gives:\[ \ln(e^{-0.3t}) = \ln(27) \]Since \( \ln(e^x) = x \), this simplifies to:\[ -0.3t = \ln(27) \]Now solve for \( t \) by dividing both sides by \(-0.3\):\[ t = \frac{\ln(27)}{-0.3} \].

Step 2: Solve Part (b)

For the equation \( e^{kt} = \frac{1}{2} \), take the natural logarithm of both sides:\[ \ln(e^{kt}) = \ln\left(\frac{1}{2}\right) \]This simplifies to:\[ kt = \ln\left(\frac{1}{2}\right) \]Solve for \( t \) by dividing both sides by \( k \):\[ t = \frac{\ln\left(\frac{1}{2}\right)}{k} \].

Step 3: Solve Part (c)

For the equation \( e^{(\ln 0.2)t} = 0.4 \), take the natural logarithm of both sides:\[ \ln(e^{(\ln 0.2)t}) = \ln(0.4) \]This simplifies to:\[ (\ln 0.2)t = \ln(0.4) \]Solve for \( t \) by dividing both sides by \( \ln 0.2 \):\[ t = \frac{\ln(0.4)}{\ln 0.2} \].

Key Concepts

Natural Logarithms

Natural logarithms are powerful tools in mathematics, especially when dealing with exponential equations. They are based on the number \(e\), which is approximately 2.718. In these equations, \(e\) is the base of the exponential function.Taking the natural logarithm (often written as \(\ln\)) helps to solve the equation by eliminating the exponential function. For example, if we have \(e^x\), taking the natural logarithm gives us \(\ln(e^x) = x\). This is because the natural logarithm is the inverse operation of exponentiation with base \(e\).
In the context of solving exponential equations like \(e^{-0.3t} = 27\), we apply the natural logarithm to both sides, allowing us to neutralize the exponential and solve for the variable, in this case, \(t\). Using properties of logarithms simplifies calculations, making natural logarithms essential when working with such equations.

Solving Equations

Solving exponential equations involves finding the unknown value that makes the equation true. This often requires using natural logarithms, as they can convert the exponential form into a simpler linear form. Let's break down the process using one of our examples:

• Consider the equation \(e^{-0.3t} = 27\).
• Take natural logarithms of both sides: \(\ln(e^{-0.3t}) = \ln(27)\).
• Simplify using the property \(\ln(e^x) = x\): \(-0.3t = \ln(27)\).
• To solve for \(t\), divide both sides by \(-0.3\): \(t = \frac{\ln(27)}{-0.3}\).

By applying this method, we can effectively solve not only single but also complex exponential equations. The concept of taking logarithms and solving the resulting linear equation is a key strategy in algebra and calculus.

Logarithmic Identities

When working with exponential equations, logarithmic identities play a crucial role. They provide shortcuts and methods for simplifying expressions, making calculations easier. Some basic logarithmic identities include:

• \(\ln(e^x) = x\) - The natural log of an exponential cancels out the \(e\).
• \(\ln(AB) = \ln A + \ln B\) - Useful when multiplying within a logarithm.
• \(\ln\left(\frac{A}{B}\right) = \ln A - \ln B\) - Helpful for dividing within a logarithm.

Let's see how these identities simplify our work. In part (c) of our example, \(e^{(\ln 0.2)t} = 0.4\), we use \(\ln(e^x) = x\) again, simplifying the left side to just \(\ln0.2) \times t\). This allows us to rearrange the equation and solve for \(t\) by using division. Overall, mastering these identities is essential to efficiently solve and manipulate exponential equations.