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Express the given quantity as a power of \(e .\) $$ \begin{array}{ll}{\text { (a) } \pi^{-x}} & {\text { (b) } x^{2 x}, x>0}\end{array} $$

Short Answer

Expert verified
Part (a): \(e^{-x\ln(\pi)}\); Part (b): \(e^{2x\ln(x)}\).

Step by step solution

01

Express the Base as Exponential with Base e (Part a)

We start with the expression \(\pi^{-x}\). To express this in terms of \(e\), we need to rewrite the base using the exponential function. We can use the logarithmic identity: \( a^b = e^{b\ln(a)} \). So for \(\pi^{-x}\), it becomes:\[\pi^{-x} = e^{-x\ln(\pi)}\]
02

Express the Base as Exponential with Base e (Part b)

Next, we have the expression \(x^{2x}\). Again, we use the identity \(a^b = e^{b\ln(a)}\) to rewrite it. Thus, \(x^{2x}\) becomes:\[x^{2x} = e^{2x\ln(x)}\]
03

Final Expression

Now we have expressed both given quantities in terms of \(e\). The expressions are:- Part (a): \(\pi^{-x} = e^{-x\ln(\pi)}\)- Part (b): \(x^{2x} = e^{2x\ln(x)}\)

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Key Concepts

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

Logarithmic Identity
The logarithmic identity is a very handy tool when you want to express a number as a power of another number. In our exercises, we used the identity
  • \(a^b = e^{b\ln(a)}\)
This means that any number raised to a power can be represented as \(e\) raised to the power of the same exponent times the natural logarithm of the base. The natural logarithm, represented by \(\ln\), is a logarithm to the base \(e\), where \(e\) is approximately 2.718.
This identity lets us transform expressions into something known as the exponential form, which is often more useful in calculus and complex analyses.

In the exercise, we used this identity to convert \(\pi^{-x}\) into \( e^{-x\ln(\pi)} \) and \(x^{2x} \) into \(e^{2x\ln(x)} \). This allows for easier computation and integration in mathematical operations.
Power of e
The number \(e\) is a fundamental constant in mathematics, much like \(\pi\). But what makes \(e\) special is its natural appearance in exponential growth and decay models, and in calculus.
  • The expression \(e^x\) represents an exponential function with \(e\) as the base.
Using \(e\) in our expressions, like the solutions \(e^{-x\ln(\pi)}\) and \(e^{2x\ln(x)}\), helps connect different mathematical operations because the derivative and integral of \(e^x\) are both quite straightforward.

Moreover, the use of \(e\) in these transformations allows for simplified equations when solving problems involving exponential growth or decay, which frequently arise in real-world scenarios, such as population growth, radioactive decay, and interest calculations.

In the context of the given exercise, expressing these powers using \(e\) facilitated their transformation into exponential functions, which are simpler to differentiate or integrate.
Base Change
Changing the base of an expression is a process widely used in mathematics to simplify expressions or adapt them for certain computations.
  • In logarithmic transformations, an expression like \(a^b\) is transformed to \(e^{b\ln(a)}\) using the identity mentioned earlier.

This base change is valuable because any exponential function can be transformed into a linear function via logarithms, making it easier to work with under various conditions.
It lets you apply logarithmic rules that might not be evident with the original base, such as simplifying expressions or solving equations.

In practice, as we did in this exercise, we transformed expressions like \(\pi^{-x}\) and \(x^{2x}\) using \(e\) as the base. This standardizes the form of all powers, making it easier to manipulate them analytically, particularly within calculus, where the use of the natural base \(e\) often simplifies derivatives and integrals.

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Most popular questions from this chapter

Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form $$V=V_{p} \sin (2 \pi f t)$$ (see the accompanying figure). In this equation, \(V_{p}\) is called the peak voltage or amplitude of the current, \(f\) is called its frequency, and \(1 / f\) is called its period. The voltages \(V\) and \(V_{p}\) are measured in volts \((\mathrm{V}),\) the time \(t\) is measured in seconds (s), and the frequency is measured in hertz (Hz). \((1 \mathrm{Hz}=1\) cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the \(r m s\) or root-mean-square value of \(V .\) By definition, this is the square root of the average value of \(V^{2}\) over one period. (a) Show that $$V_{\mathrm{rms}}=\frac{V_{p}}{\sqrt{2}}$$ [Hint: Compute the average over the cycle from \(t=0\) to \(t=1 / f,\) and use the identity \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) to help evaluate the integral. \(]\) (b) In the United States, electrical outlets supply alternating current with an rms voltage of \(120 \mathrm{V}\) at a frequency of \(60 \mathrm{Hz}\). What is the peak voltage at such an outlet?

A particle moves with a velocity of \(v(t)=3 \cos 2 t \mathrm{m} / \mathrm{s}\) along an \(s\) -axis. Find the distance traveled by the particle over the time interval \(0 \leq t \leq \pi / 8\).

Define \(F(x)\) by $$ F(x)=\int_{1}^{x}\left(3 t^{2}-3\right) d t $$ (a) Use Part 2 of the Fundamental Theorem of Calculus to find \(F^{\prime}(x) .\) (b) Check the result in part (a) by first integrating and then differentiating.

Evaluate the integrals using Part 1 of the Fundamental Theorem of Calculus. $$ \int_{-1}^{1} \frac{d x}{1+x^{2}} $$

Evaluate the integrals by any method. $$ \int_{0}^{1} \frac{x}{\sqrt{4-3 x^{4}}} d x $$

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