/*! 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 10 Evaluate each exponential expres... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 10

Evaluate each exponential expression without using a calculator. $$2^{5}$$

Short Answer

Expert verified
32

Step by step solution

01

Identify the Base and Exponent

In the expression \(2^5\), the base is 2 and the exponent is 5. This means you need to multiply the base (2) by itself as many times as the exponent indicates.
02

Expand the Expression

Expand \(2^5\) to show all multiplications: \(2 \times 2 \times 2 \times 2 \times 2\).
03

Multiply Step-by-Step

Begin multiplying the numbers step-by-step:1. \(2 \times 2 = 4\)2. \(4 \times 2 = 8\)3. \(8 \times 2 = 16\)4. \(16 \times 2 = 32\)
04

Write the Final Answer

After performing all the multiplications, you get \(32\).

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 91影视!

Key Concepts

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

headline of the respective core concept
In order to understand exponential expressions better, let's break down the key components, starting with the base and exponent. Every exponential expression has two main parts: the number being multiplied and how many times it's multiplied.
base and exponent
The base is the number that gets multiplied. It's the main number in the expression. For example, in the expression \(2^5\), the base is 2. The exponent, on the other hand, tells us how many times to multiply the base by itself. Here, the exponent is 5. This means we will be multiplying the base (2) by itself five times.
multiplication
Understanding how to carry out the multiplication step-by-step is crucial. We'll start by expanding the expression according to the exponent. For our example \(2^5\), we expand it to \(2 \times 2 \times 2 \times 2 \times 2\). Once expanded, you can begin multiplying these values one step at a time. Here's how that process looks:
  • First, multiply the first two 2's: \(2 \times 2 = 4\)
  • Next, multiply the result by the next 2: \(4 \times 2 = 8\)
  • Then, continue: \(8 \times 2 = 16\)
  • Finally, complete it: \(16 \times 2 = 32\)
This step-by-step method ensures accuracy and helps you see how the multiplication builds up from one step to the next.
step-by-step evaluation
Using a structured step-by-step approach can make evaluating exponential expressions much simpler. Here鈥檚 a full breakdown using our example \(2^5\):
  • Step 1: Identify Base and Exponent - Base: 2, Exponent: 5.
  • Step 2: Expand the Expression - Expand \(2^5\) to \(2 \times 2 \times 2 \times 2 \times 2\).
  • Step 3: Multiply Step-by-Step - Multiply each pair sequentially:
    • \(2 \times 2 = 4\)
    • \(4 \times 2 = 8\)
    • \(8 \times 2 = 16\)
    • \(16 \times 2 = 32\)
  • Step 4: Write the Final Answer - After completing the multiplications, you see that \(2^5 = 32\).
By breaking it down into manageable steps, you can follow the process easily and understand the connection between each part of the calculation.

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

Solve \(\log _{x}(9.8)=2.4 .\) Round to four decimal places.

Solve each problem. Solve the equation \(A=P e^{n t}\) for \(r,\) then find the rate at which a deposit of \(\$ 1000\) would double in 3 years compounded continuously.

Solve each problem. When needed, use 365 days per year and 30 days per month. Radioactive Decay The number of grams of a certain radioactive substance present at time \(t\) is given by the formula \(A=200 e^{-0.001 t},\) where \(t\) is the number of years. How many grams are present at time \(t=0 ?\) How many grams are present at time \(t=500 ?\)

Which exponential and logarithmic functions are increasing? Decreasing? Is the inverse of an increasing function increasing or decreasing? Is the inverse of a decreasing function increasing or decreasing? Explain.

Challenger Disaster Using data on O-ring damage from 24 previous space shuttle launches, Professor Edward R. Tufte of Yale University concluded that the number of O-rings damaged per launch is an exponential function of the temperature at the time of the launch. If NASA had used a model such as \(n=644 e^{-0.15 t},\) where \(t\) is the Fahrenheit temperature at the time of launch and \(n\) is the number of O-rings damaged, then the tragic end to the flight of the space shuttle Challenger might have been avoided. Using this model, find the number of O-rings that would be expected to fail at \(31^{\circ} \mathrm{F}\) the temperature at the time of the Challenger launch on January \(28,1986\) (Graph can't copy)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation
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.