/*! 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 28 Use properties of exponents to s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use properties of exponents to simplify each expression. First express the answer in exponential form. Then evaluate the expression. \(2^{-3} \cdot 2\)

Short Answer

Expert verified
\The simplified expression in exponential form is \(2^{-2}\) and its numerical value is \(1/4\).

Step by step solution

01

Analyze the Expression

We start with the original expression \(2^{-3} \cdot 2\). Here, we have two terms, \(2^{-3}\) and \(2\), which are being multiplied together. Both terms have the same base, 2. The first term has an exponent of -3 and the second term does not have an exponent hence it assumes the default exponent of 1.
02

Apply the Multiplication Rule for Exponents

When we multiply terms with the same base, we add the exponents. Therefore, we add the exponent -3 from the first term to the exponent 1 from the second term. That gives us the simplified expression with an exponent of \(-3 + 1 = -2\). Therefore, the expression simplifies to \(2^{-2}\).
03

Evaluate the Simplified Expression

The simplified expression is \(2^{-2}\). A negative exponent means that we take the reciprocal of the base raised to the absolute value of the exponent. Therefore, \(2^{-2}\) is equivalent to \(1/(2^2)\), which simplifies to \(1/4\).

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!

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

Write the first six terms of the geometric sequence with the first term, \(a_{1}\), and common ratio, \(r\). \(a_{1}=\frac{1}{5}, r=\frac{1}{2}\)

What is the common ratio in a geometric sequence?

You will develop geometric sequences that model the population growth for California and Texas, the two most populated U.S. states. The table shows the population of Texas for 2000 and 2010 , with estimates given by the U.S. Census Bureau for 2001 through \(2009 .\) $$ \begin{array}{|l|l|l|l|l|l|l|} \hline \text { Year } & \mathbf{2 0 0 0} & \mathbf{2 0 0 1} & \mathbf{2 0 0 2} & \mathbf{2 0 0 3} & \mathbf{2 0 0 4} & \mathbf{2 0 0 5} \\ \hline \begin{array}{l} \text { Population } \\ \text { in millions } \end{array} & 20.85 & 21.27 & 21.70 & 22.13 & 22.57 & 23.02 \\ \hline \end{array} $$ $$ \begin{array}{|l|c|c|c|c|c|} \hline \text { Year } & \mathbf{2 0 0 6} & \mathbf{2 0 0 7} & \mathbf{2 0 0 8} & \mathbf{2 0 0 9} & \mathbf{2 0 1 0} \\ \hline \begin{array}{l} \text { Population } \\ \text { in millions } \end{array} & 23.48 & 23.95 & 24.43 & 24.92 & 25.15 \\ \hline \end{array} $$ a. Divide the population for each year by the population in the preceding year. Round to two decimal places and show that Texas has a population increase that is approximately geometric. b. Write the general term of the geometric sequence modeling Texas's population, in millions, \(n\) years after \(1999 .\) c. Use your model from part (b) to project Texas's population, in millions, for the year 2020 . Round to two decimal places.

Write the first six terms of the geometric sequence with the first term, \(a_{1}\), and common ratio, \(r\). \(a_{1}=-4, r=-2\)

Write the first six terms of the geometric sequence with the first term, \(a_{1}\), and common ratio, \(r\). \(a_{1}=2, r=0.1\)

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.