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Chapter 5: Problem 83

Solve each equation and solve for \(x\) in terms of the other letters. $$y=a /\left(1+b e^{-k x}\right)$$

Short Answer

Expert verified
\(x = -\frac{1}{k} \ln\left( \frac{a - y}{yb} \right)\)

Step by step solution

01

Isolate the denominator

The given equation is \( y = \frac{a}{1 + b e^{-kx}} \). First, we need to get rid of the fraction by bringing \(1 + b e^{-kx}\) to the left side. Multiply both sides by \(1 + b e^{-kx}\) to remove the denominator: \[ y(1 + b e^{-kx}) = a \]
02

Expand and rearrange

Expand the equation: \[ y + yb e^{-kx} = a \].Next, we move the \(y\) term to the other side to isolate the exponential term: \[ yb e^{-kx} = a - y \].
03

Solve for the exponential term

Divide both sides by \(yb\) to solve for the exponential term: \[ e^{-kx} = \frac{a - y}{yb} \].
04

Apply the natural logarithm

To eliminate the exponent, take the natural logarithm of both sides: \[ -kx = \ln\left( \frac{a - y}{yb} \right) \].
05

Solve for x

Divide both sides by \(-k\) to solve for \(x\): \[ x = -\frac{1}{k} \ln\left( \frac{a - y}{yb} \right) \].

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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 variables appear as exponents. These types of equations can be tricky to solve because the variable is not linear. They often take the form of expressions with exponential terms such as \( b^x \) or \( e^{x} \). To solve an exponential equation, one common strategy involves manipulating the equation to isolate the exponential part.
  • The equation \( y = \frac{a}{1 + b e^{-kx}} \) given in our problem is a good example of an equation where \( x \) appears in an exponent.
  • This equation requires careful handling and the use of logarithms to extract the variable from its position within the exponent.
  • By reformulating an exponential equation, we can gain insights into the conditions under which the variables interact, making the task of finding solutions easier.
Breaking these equations down step-by-step is crucial for understanding how to isolate variables and solve for them.
Solving for Variables
Solving for variables involves isolating the variable of interest on one side of the equation while keeping the equation balanced. This process can require a series of manipulations depending on how the variable is embedded within the equation.
  • Start by identifying the fraction or product associated with the variable and systematically working to isolate this part.
  • In the solved problem, the first step was to clear the fraction by multiplying both sides by \(1 + b e^{-kx}\), thus simplifying the expression for further manipulation.
  • Next, we rearranged the equation to focus on the exponential term by moving other terms to the opposite side.
It's important to go through each step logically to avoid mistakes and ensure the correct isolation of the variable. Keeping track of signs and correctly applying inverse operations like division and multiplication are vital.
Natural Logarithms
Natural logarithms are a powerful tool to deal with exponential equations. The natural logarithm, denoted as \(\ln\), is the inverse function of the exponential function involving the base \(e\). This property makes natural logarithms particularly useful for solving equations where the variable is in the exponent.
  • In our step-by-step solution, when we reached \( e^{-kx} = \frac{a - y}{yb} \), taking the natural logarithm on both sides allowed us to "bring down" the exponent.
  • This application of the logarithm transforms the equation into a linear one: \( -kx = \ln\left( \frac{a - y}{yb} \right) \).
  • By applying the natural logarithm and using its properties, we made it possible to solve for \(x\) by getting rid of the exponent, which simplified the equation into a format where standard algebraic techniques could be employed.
Understanding natural logarithms and their properties can significantly simplify working with exponential equations, providing a clear path to solving for the variable.

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Most popular questions from this chapter

Prove that \(\left(\log _{a} x\right) /\left(\log _{a b} x\right)=1+\log _{a} b\)

Decide which of the following properties apply to each function. (More than one property may apply to a function.)A. The function is increasing for \(-\infty

The age of some rocks can be estimated by measuring the ratio of the amounts of certain chemical elements within the rock. The method known as the rubidium-strontium method will be discussed here. This method has been used in dating the moon rocks brought back on the Apollo missions. Rubidium-87 is a radioactive substance with a half-life of \(4.7 \times 10^{10}\) years. Rubidium- 87 decays into the substance strontium- \(87,\) which is stable (nonradioactive). We are going to derive the following formula for the age of a rock: $$T=\frac{\ln \left[\left(\mathcal{N}_{s} / \mathcal{N}_{r}\right)+1\right]}{-k}$$ where \(T\) is the age of the rock, \(k\) is the decay constant for rubidium-87, \(\mathcal{N}_{s}\) is the number of atoms of strontium-87 now present in the rock, and \(\mathcal{N},\) is the number of atoms of rubidium-87 now present in the rock. (a) Assume that initially, when the rock was formed, there were \(\mathcal{N}_{0}\) atoms of rubidium-87 and none of strontium-87. Then, as time goes by, some of the rubidium atoms decay into strontium atoms, but the to tal number of atoms must still be \(\mathcal{N}_{0} .\) Thus, after \(T\) years, we have \(\mathcal{N}_{0}=\mathcal{N}_{r}+\mathcal{N}_{s}\) or, equivalently, $$ \mathcal{N}_{s}=\mathcal{N}_{0}-\mathcal{N}_{r}$$However, according to the law of exponential decay for the rubidium-87, we must have \(\mathcal{N}_{r}=\mathcal{N}_{0} e^{k T} .\) Solve this equation for \(\mathcal{N}_{0}\) and then use the result to eliminate \(\mathcal{N}_{0}\) from equation \((1) .\) Show that the result can be written $$\mathcal{N}_{s}=\mathcal{N}_{r} e^{-k T}-\mathcal{N}_{r}$$ (b) Solve equation (2) for \(T\) to obtain the formula given at the beginning of this exercise.

Estimate a value for \(x\) such that \(\log _{2} x=100 .\) Use the approximation \(10^{3}=2^{10}\) to express your answer as a power of 10 Answer: \(10^{30}\)

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$\frac{2}{3}\left(1-e^{-x}\right) \leq-3$$
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