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Chapter 6: Problem 42

In Exercises \(40-45,\) use your ealculator to help you solve the equation or inequality. $$ e^{\sqrt{x}}=x+1 $$

Short Answer

Expert verified
Use a graphing calculator to plot and find the intersection of \( e^{\sqrt{x}} \) and \( x + 1 \).

Step by step solution

01

Understand the Equation

We start with the equation \( e^{\sqrt{x}} = x + 1 \). Here, \( e \) is the base of the natural logarithm, an irrational constant approximately equal to 2.718. Our task is to find values of \( x \) that satisfy this equation using a graphing calculator.
02

Graph the Functions

On a graphing calculator, plot the two functions \( y_1 = e^{\sqrt{x}} \) and \( y_2 = x + 1 \). The solution to our equation corresponds to the points where these two graphs intersect.
03

Find the Intersection

Use the graphing calculator's 'intersect' feature to find the values of \( x \) where the graphs intersect. This feature allows the calculator to calculate the x-value of the intersection point(s) of \( y_1 \) and \( y_2 \).
04

Verify the Solution

After finding the intersection point, substitute the value of \( x \) back into the original equation to ensure both sides equal. This confirms the accuracy of the graphically obtained solution.

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Key Concepts

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

Exponential Equations
Exponential equations involve variables appearing as exponents. In such equations, we deal with expressions like \( e^{\sqrt{x}} \). The constant \( e \) is known as Euler's Number, approximately 2.718, and is the base of natural logarithms.
These equations can sometimes be tricky since the variable is in the exponent, rather than in a typical linear or polynomial setting.
Here's a quick guide to handling exponential equations:
  • Identify whether the equation can be rearranged or simplified.
  • Look at methods such as taking the logarithm, which can help isolate the variable.
  • Understand that solutions might require numerical methods or graphing when exact algebraic manipulation is tough.
In our example, the equation \( e^{\sqrt{x}} = x + 1 \) is perfect for demonstrating how we can use graphing to find solutions, as it combines both an exponential and a linear expression.
Intersection Points
Intersection points in mathematics refer to points where two lines or curves meet 鈥� they "intersect". Finding these for the functions \( y_1 = e^{\sqrt{x}} \) and \( y_2 = x + 1 \) is central to solving our problem.
Intersection points signify the common solutions for the equations represented by the graphs. Here's why they matter in graphing solutions:
  • They represent the x-values where both expressions are equal.
  • Finding them is crucial when dealing with equations that are difficult to solve algebraically.
  • Graphing calculators have a specific feature to pinpoint these, making it easier and more precise for students.
Understanding intersection points helps tremendously in navigating problems where both terms of the equation are complex.
Graphical Solution Techniques
Graphical solution techniques refer to solving equations by observing where graphs intersect. This hands-on approach is especially useful for equations not readily solvable by algebra alone. When solving \( e^{\sqrt{x}} = x + 1 \), graphing each side individually helps visualize their interaction.
Steps for a graphical approach include:
  • Graph the two sides of the equation as separate functions, here \( y_1 \) and \( y_2 \).
  • Use the calculator's intersection feature to find where the graphs meet.
  • Interpret intersection points as solutions for the variable.
This method provides a visual insight into solution spaces and helps in cases where solutions aren't easily attainable through algebra alone.
Natural Logarithm
The natural logarithm, denoted \( \ln \), is the logarithm to the base \( e \). It's an essential tool in solving exponential equations, as it helps simplify expressions like \( e^{\sqrt{x}} = x + 1 \).
Here's how it plays into solving such equations:
  • Applying \( \ln \) can "bring down" exponents, turning them into products 鈥� this simplifies finding solutions.
  • It converts exponential growth back into manageable linear terms, which are much easier to manipulate.
  • While graphing is often used in conjunction with \( \ln \) in practice, understanding its concept is key to grasping the deeper mechanics of exponential equations.
By mastering natural logarithms, students unlock a powerful method for tackling complex equations where variables are exponents.

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Most popular questions from this chapter

The Gross Domestic Product (GDP) of the US (in billions of dollars) \(t\) years after the year 2000 can be modeled by: $$ G(t)=9743.77 e^{0.0514 t} $$ (a) Find and interpret \(G(0)\). (b) According to the model, what should have been the GDP in \(2007 ?\) In \(2010 ?\) (According to the US Department of Commerce, the 2007 GDP was \(\$ 14,369.1\) billion and the 2010 GDP was \(\$ 14,657.8\) billion.)

Use the properties of logarithms to write the expression as a single logarithm. $$ \log (x)-\frac{1}{3} \log (z)+\frac{1}{2} \log (y) $$

For each of the scenarios given in Exercises \(1-6\), \- Find the amount \(A\) in the account as a function of the term of the investment \(t\) in years. \- Determine how much is in the account after 5 years, 10 years, 30 years and 35 years. Round your answers to the nearest cent. \- Determine how long will it take for the initial investment to double. Round your answer to the nearest year. \- Find and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year. Round your answer to two decimal places. $$\$ 1000$$ is invested in an account which offers \(1.25 \%\), compounded continuously.

A finance company offers a promotion on \(\$ 5000\) loans. The borrower does not have to make any payments for the first three years, however interest will continue to be charged to the loan at \(29.9 \%\) compounded continuously. What amount will be due at the end of the three year period, assuming no payments are made? If the promotion is extended an additional three years, and no payments are made, what amount would be due?

Solve the equation analytically. $$ \ln (x+1)-\ln (x)=3 $$
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