/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Find the exact value of each log... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the exact value of each logarithm. $$ \ln \sqrt[5]{e} $$

Short Answer

Expert verified
The exact value of \(\ln \sqrt[5]{e}\) is \(\frac{1}{5}\).

Step by step solution

01

Understand the Problem

We need to find the natural logarithm of a fifth root expression, specifically \(\ln \sqrt[5]{e}\). This involves simplifying the expression to make it easier to evaluate.
02

Convert the Fifth Root to an Exponential

Recall that taking the fifth root of a number is the same as raising that number to the power of \(\frac{1}{5}\). Thus, \(\sqrt[5]{e}\) can be rewritten as \(e^{\frac{1}{5}}\).
03

Apply the Logarithmic Identity

Use the identity \(\ln(a^b) = b \cdot \ln(a)\). Substitute \(a = e\) and \(b = \frac{1}{5}\). This gives us \(\ln(e^{\frac{1}{5}}) = \frac{1}{5} \cdot \ln(e)\).
04

Evaluate the Logarithm

Recall that \(\ln(e) = 1\). Therefore, \(\frac{1}{5} \cdot \ln(e) = \frac{1}{5} \cdot 1 = \frac{1}{5}\). Thus, \(\ln \sqrt[5]{e} = \frac{1}{5}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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. The general form is \( f(x) = a^x \), where \( a \) is a positive constant. In our case, \( e \) is a common base, where \( e \) (approximately 2.718) is a natural number known as Euler's number. This base is special because it arises in many areas of mathematics, notably in calculus, where it serves as the base for natural logarithms.

An important property of exponential functions is that they describe continuous growth or decay, often used in real-world applications like population models or compound interest. When you see \( e^{\frac{1}{5}} \), it signifies a growth (or scaling) factor of \( e \) raised to the power of \( \frac{1}{5} \).
  • The function remains exponentially increasing, meaning it grows as \( x \) becomes larger.
  • It is continuously differentiable, indicating smooth transitions.
Understanding exponential expressions is key in solving problems involving powers and growth rates.
Logarithmic Identity
Logarithmic identities help simplify expressions involving logarithms. One fundamental identity is \( \ln(a^b) = b \cdot \ln(a) \), which is a direct application of logarithms' log power rule. This identity is extremely useful when dealing with powers in logarithmic terms. It simplistically means that the exponent in the logarithm can be brought forward as a multiplicative factor.

To apply this rule correctly:
  • Identify your base \( a \) and exponent \( b \). In our example, \( a = e \) and \( b = \frac{1}{5} \).
  • Rewrite the logarithm using the identity: \( \ln(e^{\frac{1}{5}}) = \frac{1}{5} \cdot \ln(e) \).
This simplification step is powerful as it reduces complex expressions into more manageable forms. In practice, you often encounter expressions where bringing down the power makes the evaluation straightforward, as with numeric calculations and algebraic manipulations.
Fifth Root
The fifth root of a number, shown as \( \sqrt[5]{x} \), is the number that, when multiplied by itself five times, equals \( x \). It can also be expressed with an exponent as \( x^{\frac{1}{5}} \).

The concept of a root can sometimes seem abstract, but in essence, it is about scaling a number back through a series of multiplications. If you imagine squaring a number as growing it by a power of 2, finding a root is the inverse operation, effectively reducing it.

Key points about the fifth root include:
  • It is part of radical expressions, like other roots (square root, cube root).
  • Helps solve polynomial equations where a number is raised to a fifth power.
  • Essential in contexts involving roots, especially in studying both real and complex numbers.
Understanding that \( \sqrt[5]{e} \) translates to \( e^{\frac{1}{5}} \) is critical in converting radical expressions into their exponential form, aiding in further simplification and computation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Graph each logarithmic function. $$ y=\log _{3} x $$

Simplify each rational expression. $$ \frac{x-5}{5-x} $$

The formula \(y=y_{0} e^{k t}\) gives the population size y of a population that experiences a relative growth rate \(k(k\) is positive if growth is increasing and \(k\) is negative if growth is decreasing). In this formula, \(t\) is time in years and \(y_{0}\) is the initial population at time \(0 .\) Use this formula to solve Exercises 55 and \(56 .\) Round answers to the nearest year. (Source for data: U.S. Census Bureau and Federal Reserve Bank of Chicago) In \(2009,\) the population of Michigan was approximately 9,970,000 and decreasing according to the formula \(y=y_{0} e^{-0.003 t}\). Assume that the population continues to decrease according to the given formula and predict how many years after which the population of Michigan will be \(9,500,000 .\) (Hint: Let \(y_{0}=9,970,000 ; y=9,500,000\), and solve for \(t\).)

The formula \(P=14.7 e^{-0.21 x}\) gives the average atmospheric pressure \(P\), in pounds per square inch, at an altitude \(x\), in miles above sea level. Use this formula to solve. Round to the nearest tenth. Find the elevation of a remote Himalayan peak if the atmospheric pressure atop the peak is 6.5 pounds per square inch.

The formula \(P=14.7 e^{-0.21 x}\) gives the average atmospheric pressure \(P\), in pounds per square inch, at an altitude \(x\), in miles above sea level. Use this formula to solve Exercises 43 through 46. Round to the nearest tenth. Find the average atmospheric pressure of Denver, which is 1 mile above sea level.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.