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Chapter 5: Problem 75

$$\text { Solve each formula for the indicated variable.}$$ $$y=A+B\left(1-e^{-c x}\right), \text { for } x$$

Short Answer

Expert verified
\( x = -\frac{1}{c} \ln\left(1 - \frac{y - A}{B}\right) \)

Step by step solution

01

Isolate the Exponential Term

The formula is given as \( y = A + B(1-e^{-cx}) \). To isolate the exponential term, subtract \( A \) from both sides: \( y - A = B(1-e^{-cx}) \).
02

Solve for the Exponential Expression

Divide both sides of the equation by \( B \) to isolate the term containing the exponential: \( \frac{y - A}{B} = 1 - e^{-cx} \).
03

Rearrange to Solve for the Exponential

Subtract 1 from both sides to further isolate the exponential term: \( e^{-cx} = 1 - \frac{y - A}{B} \).
04

Apply Natural Logarithm

Take the natural logarithm of both sides to solve for \( -cx \): \( \ln(e^{-cx}) = \ln\left(1 - \frac{y - A}{B}\right) \). This simplifies to \( -cx = \ln\left(1 - \frac{y - A}{B}\right) \).
05

Solve for \( x \)

Divide both sides by \( -c \) to isolate \( x \): \( x = -\frac{1}{c} \ln\left(1 - \frac{y - A}{B}\right) \).

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Key Concepts

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

Exponential Functions
Exponential functions play a crucial role in various mathematical equations, especially when describing growth or decay processes. An exponential function can be identified by its structure, which often includes a constant raised to the power of a variable. This function is represented as \( f(x) = a e^{bx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718.
Key properties of exponential functions include:
  • Growth or Decay: Depending on whether the exponent is positive or negative, the function can model exponential growth or decay.

  • Asymptotic Behavior: As \( x \to \infty \), the function approaches infinity or zero, depending on the exponent.

  • Continuity: Exponential functions are continuous and smooth, useful in real-world applications like population growth and radioactive decay.
Understanding these properties helps in solving equations where exponential terms need to be isolated and manipulated.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.718 \). It is particularly useful in solving equations involving exponential functions. The natural logarithm has unique properties that simplify the manipulation of exponentials.
  • Inverse Function: The natural logarithm is the inverse of the exponential function. This means that \( \ln(e^x) = x \) and \( e^{\ln(x)} = x \).

  • Product Rule: \( \ln(xy) = \ln(x) + \ln(y) \). This rule helps in breaking down complex products into simpler sums.

  • Quotient Rule: \( \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y) \). Useful for separating complex fractions.
In the problem at hand, using the natural logarithm transforms the exponential term and allows you to solve for the variable more straightforwardly.
Variable Isolation
Variable isolation is a fundamental technique in algebra used to solve equations. The goal is to express a specific variable in terms of other known quantities. Achieving this often involves algebraic manipulations such as adding, subtracting, multiplying, or dividing both sides of an equation by a common factor.
The process typically includes:
  • Eliminating Other Terms: Move all other terms to one side of the equation to focus on the term involving the variable of interest.

  • Using Inverse Operations: Apply operations that "undo" each other to simplify the equation step by step. This can include taking the logarithm to counter an exponent.

  • Checking Solutions: Always verify that the solution satisfies the original equation.
In the exercise, isolating the variable \( x \) involves handling the exponential expression and using the natural logarithm efficiently. By carefully isolating and solving for \( x \), you ensure an accurate solution to the problem.

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Most popular questions from this chapter

Newton's law of cooling says that the rate at which an object cools is proportional to the difference \(C\) in temperature between the object and the environment around it. The temperature \(f(t)\) of the object at time t in appropriate units after being introduced into an environment with a constant temperature \(T_{0}\) is $$f(t)=T_{0}+C e^{-k t}$$ where \(C\) and \(k\) are constants. Use this result. Boiling water at \(100^{\circ} \mathrm{C}\) is placed in a freezer at \(0^{\circ} \mathrm{C}\). The temperature of the water is \(50^{\circ} \mathrm{C}\) after 24 minutes. Approximate the temperature of the water after 96 minutes.

Use the change-of-base rule to find an approximation for each logarithm. $$\log _{1 / 2} 3$$

Use the change-of-base rule to find an approximation for each logarithm. $$\log _{200} 175$$

The growth of bacteria in food products makes it necessary to date some products (such as milk) so that they will be sold and consumed before the bacterial count becomes too high. Suppose that, under certain storage conditions, the number of bacteria present in a product is $$f(t)=500 e^{0.1 t}$$ where \(t\) is time in days after packing of the product and the value of \(f(t)\) is in millions. (a) If the product cannot be safely eaten after the bacterial count reaches \(3,000,000,000,\) how long will this take? (b) If \(t=0\) corresponds to January \(1,\) what date should be placed on the product?

Use the change-of-base rule to find an approximation for each logarithm. $$\log _{5} 10$$
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