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Chapter 11: Problem 4

Solve the equations using natural logs. $$ 7000 e^{t / 45}=1200 $$

Short Answer

Expert verified
Answer: The value of t is approximately -80.54, although this indicates a potential issue in the equation or problem statement as it does not correspond to a physically realizable scenario.

Step by step solution

01

Take the natural log of both sides

Using the properties of logarithms, we'll take the natural log of both sides of the equation, allowing us to eventually isolate the t variable: $$\ln(7,000 \cdot e^{t/45}) = \ln(1,200)$$
02

Simplify using log rules

Recall that \(\ln(a \cdot b) = \ln(a) + \ln(b)\), and that \(\ln(e^{c \cdot x})=c \cdot x \). Applying these rules, we now have: $$\ln(7,000) + \ln(e^{t/45}) = \ln(1,200)$$ $$\ln(7,000) + \frac{t}{45} = \ln(1,200)$$
03

Isolate the t variable

To find the value of t, we now need to isolate it from the equation: $$\frac{t}{45} = \ln(1,200) - \ln(7,000)$$ Multiplying both sides by 45, we get: $$t = 45(\ln(1,200) - \ln(7,000))$$
04

Calculate the value of t

Using a calculator, we can compute the value of t: $$t \approx 45 \cdot (-1.791759469) = -80.54$$ Notice that we have a negative value for t, which indicates the initial exponential equation is not physically realizable as given, or there is an error in the problem statement. However, the step-by-step solution provided here demonstrates the process to apply natural logs to isolate and solve for a variable in an exponential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Equations
Exponential equations are equations in which a variable appears in the exponent. These types of equations commonly occur in scenarios involving compound interest, population growth, and radioactive decay. Understanding how to solve them is crucial for interpreting real-world data. In an exponential equation, such as \(a \cdot e^{bx} = c\), the goal is often to determine the value of the variable that satisfies the equation.
To solve exponential equations, we typically employ logarithms, especially natural logarithms when dealing with the base \(e\). This involves rewriting the equation in a form that allows easy isolation of the variable. Recognizing the structure of exponential equations helps us apply the most effective methods for finding solutions.
Properties of Logarithms
Logarithms are fundamental tools in solving exponential equations. They are the inverse operations of exponentiation, which allow us to bring down variables from exponents for simplification.
One crucial property of logarithms is \(\ln(a \cdot b) = \ln(a) + \ln(b)\). This is helpful when dealing with products inside a logarithm. Another key property is \(\ln(e^x) = x\), due to the natural log's base. These properties help us manipulate the original equation to a form where the variable is more easily isolated.
  • Use addition to separate terms when products are involved.
  • Directly bring down variables from exponents using their respective coefficients.
By understanding these properties, one can transform complex equations into simpler linear forms, making them solvable using basic algebra.
Isolation of Variables
Isolating the variable is a common strategy to solve any algebraic equation, including exponential equations. It involves rearranging the equation so that the variable of interest stands alone on one side. This process often requires a series of algebraic manipulations, using operations inversely.
For example, from the original equation \(7000 \cdot e^{t/45}=1200\), we aim to have \(t\) isolated. After applying logarithms and simplifying using logarithmic properties, we re-arrange as \(\frac{t}{45} = \ln(1,200) - \ln(7,000)\). Multiplying both sides by 45 further isolates \(t\).
  • Perform operations on both sides of the equation to maintain equality.
  • Continue simplifying until the variable stands alone.
Understanding the steps needed to isolate variables is essential, as it allows you to find unknown values needed to solve exponential problems.

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Most popular questions from this chapter

Solve the equations, first approximately, as in Example \(1,\) by filling in the given table, and th猫n to four decimal places by using logarithms. $$ \begin{aligned} &\begin{array}{l} \text { Table 11.6 } \text { Solve } 2^{x}=20 \\ \hline \end{array}\\\ &\begin{array}{c|c|c|c|c} \hline x & 4.1 & 4.2 & 4.3 & 4.4 \\ \hline 2^{x} & & & & \\ \hline \end{array} \end{aligned} $$

Problems \(40-43\) concern the Krumbein phi \((\phi)\) scale of particle size, which geologists use to classify soil and rocks, defined by the formula \({ }^{15}\) $$ \phi=-\log _{2} D $$ where \(D\) is the diameter of the particle in \(\mathrm{mm}\). Some particles of clay have diameter \(0.0035 \mathrm{~mm}\). What do they measure on the \(\phi\) scale?

Solve the equations. $$ \log x=1.172 $$

Evaluate the expressions without using a calculator. Verify your answers. Example. We have \(\log _{2} 32=5\) because \(2^{5}=32\). $$ \log _{9} 9 $$

Evaluate the expressions without using a calculator. Verify your answers. Example. We have \(\log _{2} 32=5\) because \(2^{5}=32\). $$ \log _{20} 0.05 $$
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