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Chapter 10: Problem 72

In Exercises, solve for \(x\) or \(t\). $$ \left(1+\frac{0.06}{12}\right)^{12 t}=5 $$

Short Answer

Expert verified
The solution to this problem is: \( t = \frac{\ln(5)}{12 \cdot \ln\left(1+\frac{0.06}{12}\right)} \)

Step by step solution

01

Write the equation

We start with the given equation which is \[ \left(1+\frac{0.06}{12}\right)^{12 t}=5 \]
02

Apply logarithm on both sides

To make \(t\) manageable, we apply the natural log (ln) to both sides of the equation which gives us: \[ \ln\left[\left(1+\frac{0.06}{12}\right)^{12t}\right] = \ln(5) \]
03

Simplify using the properties of logarithms

Using the property of logarithms that allows us to bring the exponent down, the equation now becomes: \[ 12t \cdot \ln\left[\left(1+\frac{0.06}{12}\right)\right] = \ln(5) \]
04

Isolate the variable \(t\)

Next, we need to isolate \(t\). To do that, divide both sides by \(12 \cdot \ln\left( 1+\frac{0.06}{12}\right)\) to get: \[ t = \frac{\ln(5)}{12 \cdot \ln\left[\left(1+\frac{0.06}{12}\right)\right]} \]

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Key Concepts

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

Logarithms
Logarithms are a powerful mathematical tool to solve exponential equations. They are the inverse operations of exponentiation. When we have an equation where a variable is an exponent, like in the given exercise, logarithms help us by transforming the equation into a form that can be solved with regular algebraic techniques.
  • Logarithms express the power to which a number (the base) must be raised to get another number.
  • Common types of logarithms include base 10 (log) and the natural logarithm (ln), which uses the base e.
In this exercise, the natural logarithm is used because it simplifies calculations when dealing with exponential growth models, often reflecting natural phenomena.
The property that allows us to "bring down" the exponent is key in transforming the equation to a linear form, making the unknown visible and solvable.
Solving Equations
Solving equations involves finding the value of the variable that makes the equation true. With exponential equations, like the one in the given exercise, the unknown variable is in the exponent, making solving a bit tricky without using additional techniques.

Steps to solve such equations include:
  • Applying the logarithm to both sides. This helps in handling the exponent more easily.
  • Using properties of logarithms, such as the product, quotient, and power rules, to simplify the equation.
  • Algebraic manipulations, including isolating the variable to solve for it.
The main goal is to transform the complex exponential form into a simpler algebraic form where the variable stands out, making it straightforward to solve.
Algebraic Manipulations
Algebraic manipulations play a crucial role in simplifying and solving equations. Once the equation is logged and simplified, it's essential to use algebraic techniques to isolate the variable. These steps include dividing both sides of the equation by coefficients or terms surrounding the unknown.

In our exercise:
  • We first apply logarithms to remove the exponent form.
  • Next, we simplify using the logarithm's property, bringing down the exponent as a multiplier.
  • Finally, by dividing each term by the accompanying logarithmic expression and constants, we isolate the variable, completing the algebraic manipulation.
These techniques are fundamental not only for solving equations like these but also for tackling a wide range of algebra problems. It's all about rearranging the equation to isolate the unknown, often the final piece needed to find a solution.

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Most popular questions from this chapter

In Exercises, use the given information to write an equation for \(y\). Confirm your result analytically by showing that the function satisfies the equation \(d y / d t=C y .\) Does the function represent exponential growth or exponential decay? $$ \frac{d y}{d t}=2 y, \quad y=10 \text { when } t=0 $$

The cumulative sales \(S\) (in thousands of units) of a new product after it has been on the market for \(t\) years are modeled by \(S=C e^{k / t}\) During the first year, 5000 units were sold. The saturation point for the market is 30,000 units. That is, the limit of \(S\) as \(t \rightarrow \infty\) is 30,000 . (a) Solve for \(C\) and \(k\) in the model. (b) How many units will be sold after 5 years? (c) Use a graphing utility to graph the sales function.

In Exercises, find \(d x / d p\) for the demand function. Interpret this rate of change when the price is \(\$ 10\). $$ x=\ln \frac{1000}{p} $$

Students in a learning theory study were given an exam and then retested monthly for 6 months with an equivalent exam. The data obtained in the study are shown in the table, where \(t\) is the time in months after the initial exam and \(s\) is the average score for the class. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline t & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline s & 84.2 & 78.4 & 72.1 & 68.5 & 67.1 & 65.3 \\ \hline \end{array} $$ (a) Use these data to find a logarithmic equation that relates \(t\) and \(s\). (b) Use a graphing utility to plot the data and graph the model. How well does the model fit the data? (c) Find the rate of change of \(s\) with respect to \(t\) when \(t=2\). Interpret the meaning in the context of the problem.

The retail sales \(S\) (in billions of dollars per year) of e-commerce companies in the United States from 1999 through 2004 are shown in the table. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline t & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline S & 14.5 & 27.8 & 34.5 & 45.0 & 56.6 & 70.9 \\ \hline \end{array} $$ The data can be modeled by \(S=-254.9+121.95 \ln t\), where \(t=9\) corresponds to 1999.
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