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Chapter 5: Problem 9

Express the given quantity as a power of \(e\). (a) \(3^{\pi}\) (b) \(2^{\sqrt{2}}\)

Short Answer

Expert verified
(a) \(3^{\pi} = e^{\pi \ln 3}\); (b) \(2^{\sqrt{2}} = e^{\sqrt{2} \ln 2}\).

Step by step solution

01

Recognize the Problem Type

We need to express a number in a form that involves the exponential function with the base of the natural logarithm, which is denoted as \(e\). The goal is to find a representation that uses powers of \(e\).
02

Use Logarithmic Identity

Recall that any number \(a^b\) can be expressed using the identity \(a^b = e^{b \ln a}\). This involves converting the base to \(e\) using the natural logarithm \(\ln\).
03

Apply Identity to Part (a)

For the expression \(3^{\pi}\), use the identity: \(3^{\pi} = e^{\pi \ln 3}\). Here, \(a = 3\) and \(b = \pi\), so we replace accordingly.
04

Simplify the Expression for (a)

The expression for \(3^{\pi}\) is now \(e^{\pi \ln 3}\). This is the equivalent expression to represent \(3^{\pi}\) as a power of \(e\).
05

Apply Identity to Part (b)

For the expression \(2^{\sqrt{2}}\), use the identity: \(2^{\sqrt{2}} = e^{\sqrt{2} \ln 2}\). Here, \(a = 2\) and \(b = \sqrt{2}\).
06

Simplify the Expression for (b)

The expression for \(2^{\sqrt{2}}\) is now \(e^{\sqrt{2} \ln 2}\). This is the final form of expressing \(2^{\sqrt{2}}\) as a power of \(e\).

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Key Concepts

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

Logarithmic Identity
A logarithmic identity is a powerful mathematical tool that helps us rewrite expressions in different forms. One such useful identity for working with exponents is the expression of any number raised to a power, in terms of the exponential function with base e. Suppose we have an expression like \(a^b\). Using the identity, we can express it as \(a^b = e^{b \ln a}\). This conversion involves two key components:
  • \(b\), which is the exponent in the original expression.
  • \(\ln a\), which is the natural logarithm of the base.
  • This identity works because exponentiation and logarithms, particularly the natural logarithm, are inverse operations.
  • By expressing the base using the natural logarithm, and using the exponential function, we can effectively rewrite and comprehend complex expressions.
Understanding this identity is crucial in simplifying expressions and solving exponential functions, as it allows us to represent complex numbers in simpler forms.
Natural Logarithm
The natural logarithm, denoted as \(\ln\), is a logarithm with the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. Natural logarithms are incredibly useful in calculus and exponential growth problems.
The natural logarithm \(\ln a\) answers the question: to what power must \(e\) be raised to get \(a\)? For example, if \(\ln 3\) equals a certain number, it means that raising \(e\) to that number will result in 3.
  • Natural logarithms simplify the understanding of growth processes, decay, and many real-world phenomena.
  • They are essential in transforming expressions, particularly when using logarithmic identities.
When expressing numbers as powers of \(e\), the natural logarithm helps bridge the original base with the base \(e\), unveiling the exponential behavior of functions.
Expression Simplification
Expression simplification is aimed at making complex expressions easier to understand and work with. When we simplify expressions using powers of \(e\), logarithmic identities and natural logarithms are typically involved.
Consider the expression \(3^{\pi}\). Using the identity \(a^b = e^{b \ln a}\), we turn it into \(e^{\pi \ln 3}\). This new expression:
  • Maintains the same value as the original.
  • Links neatly to calculus principles, where exponential functions are central.
For practical purposes like solving equations or calculus problems, expressing powers in terms of \(e\) often simplifies manipulation. Remember that:
  • Handling expressions with \(e\) encourages better integration, derivation, and solving of complex problems.
  • Working in exponential form enhances our understanding and functionality within mathematics, especially in higher-level scenarios.
Mastering expression simplification can be indispensable in both academic and real-world applications, where clear and concise representation leads to better understanding and analysis.

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Most popular questions from this chapter

(a) Prove that if \(f\) is an odd function, then $$ \int_{-a}^{a} f(x) d x=0 $$ and give a geometric explanation of this result. [Hint: One way to prove that a quantity \(q\) is zero is to show that \(q=-q .]\) (b) Prove that if \(f\) is an even function, then $$ \int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x $$ and give a geometric explanation of this result. [Hint: Split the interval of integration from \(-a\) to \(a\) into two parts at \(0 .]\)

Let \(y(t)\) denote the number of \(E\). coli cells in a container of nutrient solution \(t\) minutes after the start of an experiment. Assume that \(y(t)\) is modeled by the initial-value problem $$ \frac{d y}{d t}=(\ln 2) 2^{t / 20}, \quad y(0)=20 $$ Use this model to estimate the number of \(E\). coli cells in the container 2 hours after the start of the experiment.

Consider the sum \(\sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=\left[\begin{array}{rl} & {\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right]} \\ & +\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right]\end{array}\right.\) \(=5^{3}-1^{3}=124\) For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$ \sum_{k=2}^{20}\left(\frac{1}{k^{2}}-\frac{1}{(k-1)^{2}}\right) $$

Find a positive value of \(k\) such that the area under the graph of \(y=e^{2 x}\) over the interval \([0, k]\) is 3 square units.

Determine whether the equation is true or false. Explain your answer. $$ \int \frac{2 x}{1+x^{2}} d x=\int_{1}^{1+x^{2}} \frac{1}{t} d t+C $$
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