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Chapter 3: Problem 65

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$\frac{3000}{2+e^{2 x}}=2$$

Short Answer

Expert verified
The approximate solution to the given exponential equation is \( x \approx 3.912 \).

Step by step solution

01

Isolate the Exponential

First, we isolate the term \( e^{2x} \). To begin, multiply both sides of the original equation by the denominator of the left side to get rid of the fraction:\( 3000 = 2(2 + e^{2x}) \)Then, simplify the right side:\( 3000 = 4 + 2e^{2x} \)Lastly, subtract 4 from both sides:\( 2996 = 2e^{2x} \)
02

Further Isolate \(e^{2x}\)

Now, we divide both sides by 2:\( 1498 = e^{2x} \)
03

Convert to Logarithmic Form

Conversion to logarithmic form allows us to solve the equation. The equation\( 1498 = e^{2x} \)becomes\( \ln(1498) = 2x \)
04

Solve for the variable \(x\)

Divide both sides by 2 to solve for \(x\):\( x = \frac{\ln(1498)}{2} \)
05

Get Approximate Value

Finally, we calculate the right side:\( x \approx 3.912 \)

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Key Concepts

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

Exponential Function
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In the given exercise, the term e^{2x} is an example of an exponential function, where e (approximately 2.71828) is the base, and 2x is the variable exponent.

These functions are crucial in modeling growth and decay processes such as population growth, radioactive decay, and interest compounding. The key characteristic of an exponential function is that the rate of change is proportional to the function's current value, leading to rapid increases or decreases depending on the exponent's sign and magnitude.

Understanding how to manipulate and solve equations involving exponential functions is essential in various scientific and mathematical applications. They often require special techniques for isolating the variable, as we see in the provided solution where algebraic steps are taken to isolate e^{2x}.
Logarithmic Form
Logarithmic form is the inverse operation of exponentiation, and it is used to solve exponential equations like the one in our exercise. The logarithm, often denoted as log (for base 10) or ln (for the natural logarithm with base e), answers the question: 'To what power must the base be raised to produce a given number?'.

For the equation e^{2x} = 1498, we can rewrite it in logarithmic form as ln(1498) = 2x. The natural logarithm ln is used because our base is the natural number e. This conversion from exponential to logarithmic form is a vital step in solving for the variable, as logarithms allow us to bring the exponent down to a position where we can perform algebraic operations to solve for it.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying expressions to isolate and solve for unknown variables. When solving exponential equations, algebraic manipulation often includes operations such as multiplying or dividing both sides of the equation, distributing, and combining like terms.

In our exercise, we start by eliminating the fraction to isolate the term with e^{2x}. After simplifying and isolating e^{2x}, we apply division to both sides. These steps demonstrate the importance of algebraic manipulation in preparing an equation for conversion to the logarithmic form that ultimately allows for the variable to be solved.

Effective algebraic manipulation requires a solid grasp of basic algebraic principles, and with practice, it enables the solving of various complex equations including those involving exponential functions.

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Most popular questions from this chapter

A $$\$ 120,000$$ home mortgage for 30 years at \(7 \frac{1}{2} \%\) has a monthly payment of $$\$ 839.06 .$$ Part of the monthly payment is paid toward the interest charge on the unpaid balance, and the remainder of the payment is used to reduce the principal. The amount that is paid toward the interest is \(u=M-\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}\) and the amount that is paid toward the reduction of the principal is \(v=\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}\) In these formulas, \(P\) is the size of the mortgage, \(r\) is the interest rate, \(M\) is the monthly payment, and \(t\) is the time (in years). (a) Use a graphing utility to graph each function in the same viewing window. (The viewing window should show all 30 years of mortgage payments.) (b) In the early years of the mortgage, is the larger part of the monthly payment paid toward the interest or the principal? Approximate the time when the monthly payment is evenly divided between interest and principal reduction. (c) Repeat parts (a) and (b) for a repayment period of 20 years \((M=\$ 966.71) .\) What can you conclude?

Solve the equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility. $$2 x \ln \left(\frac{1}{x}\right)-x=0$$

The number of bacteria in a culture is increasing according to the law of exponential growth. The initial population is 250 bacteria, and the population after 10 hours is double the population after 1 hour. How many bacteria will there be after 6 hours?

Rewrite each verbal statement as an equation. Then decide whether the statement is true or false. Justify your answer. The logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers.

Complete the table for the time \(t\) (in years) necessary for \(P\) dollars to triple if interest is compounded continuously at rate \(r\). $$ \begin{array}{|l|l|l|l|l|l|l|} \hline r & 2 \% & 4 \% & 6 \% & 8 \% & 10 \% & 12 \% \\ \hline t & & & & & & \\ \hline \end{array} $$
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