/*! 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 137 Determine whether each statement... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 137

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Because the equations \(2^{x}=15\) and \(2^{x}=16\) are similar, I solved them using the same method.

Short Answer

Expert verified
Yes, the statement makes sense. Although the specific solutions are different, the same method, which involves using the properties of logarithms, is applicable to solve both given equations \(2^{x}=15\) and \(2^{x}=16\).

Step by step solution

01

Understanding the problem

The statement in question is examining if methods of solutions can be applied uniformly across similar problems. The two equations given, \(2^{x}=15\) and \(2^{x}=16\), indeed have the same structure, both being exponential equations with base 2. The similarity ends there though as the right hand side of the equations are different. This influences the exact solution but does not impact the methodology to solve them.
02

Explaining the solution method

Exponential equations like the ones given can be solved using the property of logarithms. The general process involves taking the logarithm of both sides of the equation, often base 2 logarithm when the base of the exponential is 2. This will transform the equation into a format that isolates \(x\) and makes the solution straightforward. For the equation \(2^{x}=15\), the solution will be \(x = \log_{2}15\). Similarly, for the equation \(2^{x}=16\), the solution will be \(x = \log_{2}16 = 4\).
03

Concluding the reasoning

The method to solve both equations is indeed the same, rooted in taking advantage of the properties of logarithms. The specific solution steps differ slightly between the two equations due to the different numbers on the right hand side of the equations. So the statement does make sense, because the equations are similar enough to use the same method to solve them.

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!

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

In parts (a)-(c), graph \(f\) and \(g\) in the same viewing rectangle. a. \(f(x)=\ln (3 x), g(x)=\ln 3+\ln x\) b. \(f(x)=\log \left(5 x^{2}\right), g(x)=\log 5+\log x^{2}\) c. \(f(x)=\ln \left(2 x^{3}\right), g(x)=\ln 2+\ln x^{3}\) d. Describe what you observe in parts (a)-(c). Generalize this observation by writing an equivalent expression for \(\log _{b}(M N),\) where \(M>0\) and \(N>0\) e. Complete this statement: The logarithm of a product is equal to _________.

Use the exponential decay model, \(A=A_{0} e^{k t},\) to solve Exercises \(28-31 .\) Round answers to one decimal place. The half-life of lead is 22 years. How long will it take for a sample of this substance to decay to \(80 \%\) of its original amount?

In Exercises \(53-56,\) rewrite the equation in terms of base \(e\). Express the answer in terms of a natural logarithm and then round to three decimal places. $$ y=2.5(0.7)^{x} $$

Would you prefer that your salary be modeled exponentially or logarithmically? Explain your answer.

Graph \(f\) and \(g\) in the same viewing rectangle. Then describe the relationship of the graph of g to the graph of \(f\). $$f(x)=\ln x, g(x)=\ln (x+3)$$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

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.