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Write the rule of the function in the form \(\left.f(x)=P e^{k x} . \text { (See the discussion and box after Example } 11 .\right)\) $$g(x)=-16\left(30.5^{x}\right)$$

Short Answer

Expert verified
Question: Rewrite the function \(g(x) = -16\left(30.5^{x}\right)\) in the form \(f(x) = P e^{kx}\), identifying the constant term P and the exponential term k. Answer: The function can be rewritten as \(f(x) = -16 e^{x\ln(30.5)}\), where \(P = -16\) and \(k = \ln(30.5)\).

Step by step solution

01

Identify constant term P

From the given function \(g(x) = -16(30.5^x)\), we can identify that \(P = -16\). So the function now looks like: \(f(x)=-16e^{kx}\).
02

Convert base of exponent to e

To rewrite \(30.5^x\) to base \(e\), we'll use the relationship \(a^x = e^{x\ln(a)}\). In this case, we have \(a = 30.5\). So, \(30.5^x = e^{x\ln(30.5)}\).
03

Rewrite the exponential term

Substitute the new exponential term from step 2 into the function. So, \(f(x) = -16 e^{x\ln(30.5)}\).
04

Identify k

Now, we can simplify the function and identify \(k\) as the constant multiplier for \(x\). Comparing this function with the desired form \(f(x)=Pe^{kx}\), we can see that the constant multiplier \(k\) is equal to \(\ln(30.5)\). So, the final function is: \(f(x) = -16 e^{x\ln(30.5)}\). We have successfully rewritten the given function in the form \(f(x)=Pe^{kx}\), with \(P=-16\) and \(k=ln(30.5)\).

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Key Concepts

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

Function Representation
Understanding how a function is represented is essential in mathematics. A function is a relationship between a set of inputs and outputs, often expressed in a specific form. In the context of exponential functions, a common representation is
  • \( f(x) = Pe^{kx} \),
where:
  • \( P \) represents the initial value or the coefficient, indicating the function's starting point when \( x=0 \),
  • \( e \) is the base of natural logarithms, approximately equal to 2.71828,
  • \( k \) is a constant that determines the rate at which the function grows or decays, depending on its sign.
This representation is particularly useful in modeling continuous growth or decay processes, as it uses an exponential function to depict changes over time.
The function format guides us in determining how fast a process speeds up or slows down. This is crucial in fields like population dynamics, finance, and physics, where growth and decay are prevalent.
Understanding this representation allows for accurate predictions and analyses in mathematical models.
Conversion to Base e
The conversion to base \( e \) in exponential functions is a key step for simplifying the expression and utilizing natural logarithms. Given a function in the form of \( a^x \), it may sometimes be more practical to convert it to a base of \( e \).
This is done using the property:
  • \( a^x = e^{x \ln(a)} \).
In this expression,
  • \( a \) is the original base of the exponential term,
  • \( x \) remains the exponent as before, and
  • \( \ln(a) \) is the natural logarithm of \( a \).
This transformation allows the use of exponential rules with base \( e \), which simplifies differentiation and integration in calculus.
Converting to base \( e \) can provide a more straightforward expression for various mathematical operations.
It's particularly helpful when solving differential equations or analyzing exponential growth or decay in continuous-time models.
Exponential Growth and Decay
Exponential functions are vital in representing processes of growth and decay. They model situations where the rate of change is proportional to the value of the function at any time. This comes in two main forms:
  • Growth: When \( k > 0 \), the function \( f(x) = Pe^{kx} \) describes **exponential growth**. Here, the function value increases as \( x \) increases, often seen in populations or investments compounding continuously.
  • Decay: When \( k < 0 \), the function indicates **exponential decay**. In this case, \( f(x) \) decreases over time, common in radioactive decay or cooling processes.
In both scenarios, the constant \( P \) represents the initial quantity before any growth or decay begins.
Exponential functions can drastically change over small differences in \( x \) due to their very nature, emphasizing the importance of understanding this type of mathematical model.
Utilizing the exponential growth and decay models allows us to predict and analyze real-world phenomena involving rapid changes over time, thereby aiding in planning, decision-making, and strategic forecasting.

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Most popular questions from this chapter

Look back at Section \(4.4,\) where the basic properties of graphs of polynomial functions were discussed. Then review the basic properties of the graph of \(f(x)=a^{x}\) discussed in this section. Using these various properties, give an argument to show that for any fixed positive number \(a(\neq 1),\) it is not possible to find a polynomial function \(g(x)=c_{n} x^{n}+\dots+c_{1} x+c_{0}\) such that \(a^{x}=g(x)\) for \(a l l\) numbers \(x .\) In other words, no exponential function is a polynomial function. However, see Exercise \(81 .\)

The population of India was approximately 1030 million in 2001 and was 967 million in \(1997 .\) If the population continues to grow exponentially at the same rate, what will it be in \(2010 ?\)

A harmonic sum is a sum of this form: $$ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{k} $$ (a) Compute \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}, 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5},\) and $$ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} $$ (b) How many terms do you need in a harmonic sum for it to exceed three? (c) It turns out to be hard to determine how many terms you would need for the sum to exceed 10. It will take thousands of terms, more than you would want to plug into a calculator. Using calculus, we can derive this lower-bound formula: \(\sum_{i=1}^{n} \frac{1}{i} > \ln n .\) It means that the harmonic sum with \(n\) terms is always greater than \(\ln n .\) Use this formula to find a value of \(n\) such that the harmonic sum with \(n\) terms is greater than ten. (d) Calculus also gives us an upper-bound formula: \(\sum_{i=1}^{n} \frac{1}{i} < \ln n+1 .\) Estimate the harmonic sum with 100,000 terms. How close is your estimate to the real number?

Sketch a complete graph of the function. $$g(x)=(5 / 2)^{x}$$

One person with a flu virus visited the campus. The number \(T\) of days it took for the virus to infect \(x\) people was given by: $$ T=-.93 \ln \left[\frac{7000-x}{6999 x}\right] $$ (a) How many days did it take for 6000 people to become infected? (b) After two weeks, how many people were infected?

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