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Chapter 5: Problem 8

Use the properties of exponents to simplify each expression. In Exercises 9 and \(10,\) write the answers in the form \(b^{n}\), where \(b\) and \(n\) are real numbers. $$\frac{10^{\pi+2}}{10^{\pi-2}}$$

Short Answer

Expert verified
\(10^4\)

Step by step solution

01

Identify the Base and the Exponents

In the expression \( \frac{10^{\pi+2}}{10^{\pi-2}} \), the base is 10 for both the numerator and the denominator. The exponents are \(\pi + 2\) in the numerator and \(\pi - 2\) in the denominator.
02

Apply the Quotient Rule of Exponents

The Quotient Rule states that \(\frac{a^m}{a^n} = a^{m-n}\), where \(a\) is the base, and \(m\) and \(n\) are the exponents. Applying this rule, we have: \(\frac{10^{\pi+2}}{10^{\pi-2}} = 10^{(\pi+2)-(\pi-2)}\).
03

Simplify the Exponent

Simplify the exponent by performing the subtraction: \((\pi + 2) - (\pi - 2) = \pi + 2 - \pi + 2 = 4\).
04

Rewrite the Expression in Simplified Form

Using the simplified exponent, rewrite the expression: \(10^4\). This means the original expression simplifies to \(10^4\).

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Key Concepts

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

Quotient Rule of Exponents
When dealing with exponents, the quotient rule is an essential concept. This rule tells us how to simplify expressions that involve division of like bases with exponents. Using the quotient rule,
  • If you have an expression of the form \( \frac{a^m}{a^n} \), then it simplifies to \( a^{m-n} \).
  • "Like bases" mean that both the numerator and the denominator have the same base \(a\).
  • The exponents \(m\) and \(n\) are subtracted from each other.
Starting with the expression \( \frac{10^{\pi+2}}{10^{\pi-2}} \), notice that the base of both the numerator and the denominator is 10. So, according to the quotient rule, we subtract the exponent in the denominator from the exponent in the numerator. This means that \( (\pi + 2) - (\pi - 2) \) is calculated, resulting in a much simpler expression. This step is vital and reduces complex division into a manageable form.
Simplifying Expressions
Simplifying expressions is central to many mathematical problems, and exponents are no exception. It involves reducing an expression into its simplest form, making it easier to work with or understand.
  • In the context of the exercise \( \frac{10^{\pi+2}}{10^{\pi-2}} \), we first simplify the exponent part of the expression.
  • After applying the quotient rule, the process leads us to calculate \( (\pi + 2) - (\pi - 2) \).
  • This arithmetic operation simplifies to \(4\).
Simplification involves combining or cancelling elements wherever possible, leaving you with an expression that clearly highlights the core values involved鈥攊n this case, \(10^4\). By simplifying as much as possible, you avoid unnecessary complexity in expression handling.
Exponent Rules
Understanding exponent rules is crucial to mastering exponent-related problems in mathematics. These rules provide guidelines for dealing with exponents during various operations such as multiplication, division, and raising powers to powers. The most pertinent rules include:
  • Product Rule: \(a^m \cdot a^n = a^{m+n}\). This rule is used when multiplying like bases.
  • Quotient Rule: \(\frac{a^m}{a^n} = a^{m-n}\). We use this when dividing like bases, as seen in the exercise.
  • Power of a Power Rule: \((a^m)^n = a^{m \cdot n}\). This rule applies when raising an exponent to another power, multiplying the exponents.
  • Zero Exponent Rule: \(a^0 = 1\) for any non-zero \(a\).
  • Negative Exponent Rule: \(a^{-n} = \frac{1}{a^n}\), where \(a eq 0\).
By applying these rules appropriately, complex mathematical problems become manageable and clear. Each of these rules plays a role in simplifying expressions and solving problems efficiently. Using these rules not only helps in exams but also in practical applications of mathematics.

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Most popular questions from this chapter

(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$e^{2 t+3}=10$$

The intensity of the sounds that the human ear can detect varies over a very wide range of values. For instance, a whisper from 1 meter away has an intensity of approximately \(10^{-10}\) watts per square meter \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\), whereas, from a distance of 50 meters, the intensity of a launch of the Space Shuttle is approximately \(10^{8} \mathrm{W} / \mathrm{m}^{2} .\) For a sound with intensity \(I\), the sound level \(\beta\) is defined by $$ \beta=10 \log _{10}\left(I / I_{0}\right) $$ where the constant \(I_{0}\) is the sound intensity of a barely audible sound at the threshold of hearing. The units for the sound level \(\beta\) are decibels, abbreviated dB. (a) Solve the equation \(\beta=10 \log _{10}\left(1 / I_{0}\right)\) for \(I\) by first dividing by 10 and then converting to exponential form. (b) The sound level for a power lawnmower is \(\beta=100 \mathrm{db}\). and that for a cat purring is \(\beta=10\) db. Use your result in part (a) to determine how many times more intense is the power mower sound than the cat's purring.

Simplify each expression. (a) \(\ln e\) (b) \(\ln e^{-2}\) (c) \((\ln e)^{-2}\)

The half-life of plutonium-241 is 13 years. (a) How much of an initial 2 -g sample remains after 5 years? (b) Find the time required for \(90 \%\) of the 2 -g sample to decay. Hint: If \(90 \%\) has decayed, then \(10 \%\) remains.

A function \(f\) with domain \((1, \infty)\) is defined by the equation \(f(x)=\log _{x} 2\) (a) Find a value for \(x\) such that \(f(x)=2\) (b) Is the number that you found in part (a) a fixed point of the function \(f ?\)
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