/*! 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 29 Find all the real-number roots o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 29

Find all the real-number roots of each equation. In each case, give an exact expression for the root and also (where appropriate) a calculator approximation rounded to three decimal places. $$2^{5 x}=3^{x}\left(5^{x+3}\right)$$

Short Answer

Expert verified
The real-number root is approximately \(x \approx 6.347\), with exact expression \(x = \frac{\ln 125}{5 \ln 2 - \ln 15}\).

Step by step solution

01

Understand the Equation

The given equation is \(2^{5x} = 3^x (5^{x+3})\). Our goal is to find all the real-number roots.
02

Simplify the Right Side of Equation

Rewrite the right side: \(3^x (5^{x+3}) = 3^x \cdot 5^x \cdot 5^3 = 3^x \cdot 5^x \cdot 125\).
03

Equate and Simplify Powers of x

Now, equate the expressions so: \(2^{5x} = 3^x \cdot 5^x \cdot 125\). Simplifying the right side gives \(2^{5x} = 125 \cdot (3 \cdot 5)^x = 125 \cdot 15^x\).
04

Distribute and Equate Exponents

Divide both sides by 125: \(\frac{2^{5x}}{125} = 15^x\) or equivalently, \(2^{5x} = 125 \cdot 15^x\). Now take the logarithm (let's use the natural log) of both sides: \(\ln(2^{5x}) = \ln(125 \cdot 15^x)\).
05

Apply Logarithmic Identities

Using logarithm, this becomes \(5x \ln 2 = x \ln 15 + \ln 125\).
06

Isolate x

Rearrange for x: \(5x \ln 2 - x \ln 15 = \ln 125\). Factor out x: \(x(5 \ln 2 - \ln 15) = \ln 125\).
07

Solve for x

Solve for x: \(x = \frac{\ln 125}{5 \ln 2 - \ln 15}\).
08

Calculate Exact Expression and Approximate

Using a calculator, plug in the values: \(\ln 125 \approx 4.8283\), \(5 \ln 2 \approx 3.4657\), and \(\ln 15 \approx 2.7081\). So, \(x \approx \frac{4.8283}{3.4657 - 2.7081} \approx 6.347\). This is the calculator approximation, and the exact expression is \(\frac{\ln 125}{5 \ln 2 - \ln 15}\).

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.

Real-Number Roots
Real-number roots of an equation are values of the variable that satisfy the equation, using real numbers. In this context, the numbers are not imaginary, meaning they are not complex and can be found on the continuous number line. When we're tasked with finding these roots, we first simplify the equation as much as possible.

In the given equation, we're dealing with exponential terms, which means it's important to express all the terms in a format that can easily relate to each other. This allows us to clearly see and solve for the variable.
  • Start by identifying possible transformations to match base terms.
  • Simplify the equation by breaking it down, such as factoring expressions.
The end goal is to equate the exponential parts to solve for "x", thus finding the real-number roots that satisfy the original equation.
Logarithmic Identities
Logarithmic identities are crucial for simplifying and solving exponential equations. They help us to work with expressions involving powers and roots. Key identities include the product, quotient, and power rules which allow us to transform the equation into a form that is easier to manage.

For instance, when dealing with our original equation, we use the natural logarithm to both sides like so: \[ \ln(2^{5x}) = \ln(125\cdot15^x) \] This makes use of the power property of logarithms: \[ \ln(a^b) = b\ln(a) \] By applying these identities:
  • We can break down the equation into components that are linear with respect to "x".
  • This vastly simplifies solving for the variable.
Understanding these identities allows us to transform and manipulate the equation to uncover solutions.
Calculator Approximation
Calculator approximation is useful in contexts where the exact expression is complex and a numerical answer is required. It provides a practical way to express the solution of a problem.

In our example, once the exact expression is derived:
  • We plug in the logarithmic values to a calculator.
  • This gives a decimal approximation for "x", essential for practical applications.
It鈥檚 crucial to be mindful of rounding and accuracy. In this exercise, rounding is done to three decimal places, with the approximation reaching around 6.347. This approach helps students verify their solutions and understand how equations behave under real-world mathematical conditions.
Exact Expression
Finding the exact expression of the roots in terms of known constants or operations provides a definitive solution to an equation. While calculator approximations give us a numerical value, the exact expression remains free of rounding errors, providing a pure mathematical solution.

In the exercise, we derived the exact expression for "x" as:
\[ x = \frac{\ln 125}{5 \ln 2 - \ln 15} \] This form:
  • Shows how "x" is dependent on logarithmic values or constants.
  • Ensures we have a precise algebraic representation, valuable for understanding the deeper relationships within the equation.
Emphasizing exact expressions aids in cultivating mathematical precision and resilience against miscalculation.

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 inequalities. \(\log _{\pi}\left[\log _{4}\left(x^{2}-5\right)\right]<0 \quad\) Hint: In the expression \(\log _{b} y, y\) must be positive.

Radiocarbon Dating: Because rubidium-87 decays so slowly, the technique of rubidium-strontium dating is generally considered effective only for objects older than 10 million years. In contrast, archeologists and geologists rely on the radiocarbon dating method in assigning ages ranging from 500 to 50,000 years. Two types of carbon occur naturally in our environment: carbon-12, which is nonradioactive, and carbon-14, which has a half-life of 5730 years. All living plant and animal tissue contains both types of carbon, always in the same ratio. (The ratio is one part carbon- 14 to \(10^{12}\) parts carbon-12.) As long as the plant or animal is living, this ratio is maintained. When the organism dies, however, no new carbon-14 is absorbed, and the amount of carbon-14 begins to decrease exponentially. since the amount of carbon-14 decreases exponentially, it follows that the level of radioactivity also must decrease exponentially. The formula describing this situation is $$\mathcal{N}=\mathcal{N}_{0} e^{k T}$$ where \(T\) is the age of the sample, \(\mathcal{N}\) is the present level of radioactivity (in units of disintegrations per hour per gram of carbon), and \(\mathcal{N}_{0}\) is the level of radioactivity \(T\) years ago, when the organism was alive. Given that the half-life of carbon-14 is 5730 years and that \(\mathcal{N}_{0}=920\) disintegrations per hour per gram, show that the age \(T\) of a sample is given by $$T=\frac{5730 \ln (\mathcal{N} / 920)}{\ln (1 / 2)}$$

If \(a^{2}+b^{2}=7 a b,\) where \(a\) and \(b\) are positive, show that $$\log \left[\frac{1}{3}(a+b)\right]=\frac{1}{2}(\log a+\log b)$$ no matter which base is used for the logarithms (but it understood that the same base is used throughout).

Use the half-life information to complete each table. (The formula \(\mathcal{N}=\mathcal{N}_{0} e^{k t}\) is not required.) (a) Uranium-228: half-life \(=550\) seconds$$\begin{array}{llllll}t \text { (seconds) } & 0 & 550 & 1100 & 1650 & 2200 \\\\\mathcal{N} \text { (grams) } & 8 & & & \\\\\hline\end{array}$$ (b) Uranium-238: half-life \(=4.9 \times 10^{9}\) years$$\begin{array}{lcccccc}\hline \multirow{2}{*}\begin{array}{l}\text { t (years) } \\\\\mathcal{N} \text { (grams) }\end{array} & \multicolumn{2}{c}0 \\\& \multicolumn{2}{c}10 & 5 & 2.5 & 1.25 & 0.625 \\\\\hline\end{array}$$

The Chernobyl nuclear explosion (in the former Soviet Union, on April 26,1986 ) released large amounts of radioactive substances into the atmosphere. These substances included cesium-137, iodine-131, and strontium-90. Although the radioactive material covered many countries, the actual amount and intensity of the fallout varied greatly from country to country, due to vagaries of the weather and the winds. One area that was particularly hard hit was Lapland, where heavy rainfall occurred just when the Chernobyl cloud was overhead. (a) Many of the pastures in Lapland were contaminated with cesium-137, a radioactive substance with a half- life of 33 years. If the amount of cesium- 137 was found to be ten times the normal level, how long would it take until the level returned to normal? Hint: Let \(\mathcal{N}_{0}\) be the amount that is ten times the normal level. Then you want to find the time when \(\mathcal{N}(t)=\mathcal{N}_{0} / 10\) (b) Follow part (a), but assume that the amount of cesium-137 was 100 times the normal level. Remark: Several days after the explosion, it was reported that the level of cesium- 137 in the air over Sweden was 10,000 times the normal level. Fortunately there was little or no rainfall.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

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.