/*! 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 53 Use the One-to-One Property to s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the One-to-One Property to solve the equation for \(x\). $$\left(\frac{1}{2}\right)^{x}=32$$

Short Answer

Expert verified
After analyzing and solving the problem step by step, the solution for the equation \((\frac{1}{2})^{x} = 32\) is \(x = -5\).

Step by step solution

01

Rearrange the Equation

Notice how \(32 = 2^{5}\) and also \(\frac{1}{2} = 2^{-1}\). Using these properties we can rewrite our equation so that both sides have the same base. So the equation will be written as \((2^{-1})^{x} = 2^{5}\)
02

Simplify the exponent on the left

By rules of exponents \((a^{b})^{c} = a^{b*c}\) the left-hand side simplifies as \(2^{-x} = 2^{5}\)
03

Apply the One-to-One Property of Exponents

Now we have the equation in the form \(a^{m} = a^{n}\), where we can see that \(a = 2\). Here, -x corresponds to m and 5 corresponds to n. By the One-to-One Property, when the bases are the same (in this case 2), the exponents must be equal. Hence -x = 5.
04

Solve for x

To isolate x, we just multiply both sides of the equation by -1. Therefore, x = -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.

Solving Exponential Equations
Understanding how to solve exponential equations is essential for students in mathematics, especially when dealing with growth and decay problems or interest calculations. Exponential equations have the form of \(a^{x}=b\), where \(a\) is the base, \(x\) is the exponent, and \(b\) is a constant.

To solve these types of equations, we often look to have both sides of the equation with the same base to apply the One-to-One Property of Exponents. This means, if \(a^{m}=a^{n}\), then \(m=n\), provided that \(a\) is not equal to zero. In the given exercise, we first recognize that both 32 and \(\frac{1}{2}\) can be rewritten with a base of 2. This allows us to set the exponents equal to each other and solve for \(x\).

The concept is a bit like matching puzzle pieces: if the shapes (bases) match perfectly, the other attributes (exponents) have to be identical for the puzzle to be complete.
Rules of Exponents
To simplify and manipulate exponential expressions effectively, students need to be familiar with the fundamental rules of exponents. These rules help to transform expressions into an equivalent form that is often easier to work with.

Some of these key rules include:
  • The Product Rule: \(a^{m}\times a^{n}=a^{m+n}\),
  • The Quotient Rule: \(\frac{a^{m}}{a^{n}}=a^{m-n}\),
  • The Power of a Power Rule: \(\left(a^{m}\right)^{n}=a^{m\times n}\),
  • The Power of a Product Rule: \(\left(ab\right)^{m}=a^{m}\times b^{m}\),
  • The Zero Exponent Rule: \(a^{0}=1\), provided that \(a\) is not zero.
Students should practice applying these rules to become proficient in simplifying and solving exponential equations, as they are pivotal in algebra and higher-level math courses.
Exponent Simplification
When simplifying expressions with exponents, the goal is to reduce them to a form that makes them easier to interpret or use in further calculations. Following the rules of exponents can simplify even the most intimidating expression into something much more manageable.

In the exercise, we used the exponent simplification to rewrite \(\frac{1}{2}\) as \(2^{-1}\) and used the power of a power rule to combine the exponents, turning \(\left(2^{-1}\right)^{x}\) into \(2^{-x}\). This effectively prepared us to employ the One-to-One Property and match the exponents. Working through many practice problems will help students gain confidence in their ability to simplify complex exponential expressions and solve them with ease.

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 the equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility. $$2 x \ln \left(\frac{1}{x}\right)-x=0$$

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$3-\ln x=0$$

Let \(f(x)=\log _{a} x\) and \(g(x)=a^{x},\) where \(a>1\) (a) Let \(a=1.2\) and use a graphing utility to graph the two functions in the same viewing window. What do you observe? Approximate any points of intersection of the two graphs. (b) Determine the value(s) of \(a\) for which the two graphs have one point of intersection. (c) Determine the value(s) of \(a\) for which the two graphs have two points of intersection.

Solve the equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility. $$e^{-2 x}-2 x e^{-2 x}=0$$

A conservation organization releases 100 animals of an endangered species into a game preserve. The organization believes that the preserve has a carrying capacity of 1000 animals and that the growth of the pack will be modeled by the logistic curve \(p(t)=\frac{1000}{1+9 e^{-0.1656 t}}\) where \(t\) is measured in months (see figure). (a) Estimate the population after 5 months. (b) After how many months will the population be \(500 ?\) (c) Use a graphing utility to graph the function. Use the graph to determine the horizontal asymptotes, and interpret the meaning of the asymptotes in the context of the problem.

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.