/*! 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 13 Solve the equation. \(\sqrt[3]... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 13

Solve the equation. \(\sqrt[3]{5}=5^{t}\)

Short Answer

Expert verified
t = 1/3

Step by step solution

01

Recognize the Equation Format

Identify the given equation, which is a cube root equation. The equation is \(\sqrt[3]{5} = 5^{t}\).
02

Convert Cube Root to Exponential Form

Rewrite the cube root \(\sqrt[3]{5}\) as an exponent. The cube root of 5 can be expressed as \(5^{1/3}\), so the equation becomes \(5^{1/3} = 5^{t}\).
03

Equate the Exponents

Since the bases are the same, set the exponents equal to each other. Therefore, \(1/3 = t\).
04

Solve for t

To find the value of \(t\), recognize that the equation \(t = 1/3\) represents the solution. So, \(t = 1/3\).

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Key Concepts

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

Exponential Equations
When solving exponential equations, it's important to understand the role of exponents and bases. An exponential equation is one where variables appear as exponents. For example, in the equation \(\text{5}^t = \text{5}^{1/3}\), both sides have the same base. By equating the exponents when bases are identical, we simplify the problem. Here, because the base is 5 on both sides, we can set the exponents equal to each other, leading to \(t = \frac{1}{3}\). This approach simplifies solving complex problems by converting them into more manageable forms.
Cube Roots
Understanding cube roots is crucial for solving cube root equations. The cube root of a number is the value that, when multiplied by itself three times, gives the original number. Mathematically, the cube root of a number \(\text{x}\) is written as \(\text{\textbackslash sqrt[3]\textbraceleft x \textbraceright}\). In our problem, \(\text{\textbackslash sqrt[3]\textbraceleft 5 \textbraceright}\) is the cube root of 5. This can also be expressed as \( 5^{1/3}\), since the root can be represented as an exponent of one-third. Recognizing and converting between these forms allows easier manipulation and solving of equations.
Algebraic Manipulation
Algebraic manipulation involves rewriting equations to simplify or solve them. In the given problem, we start by identifying that we have a cube root equation, \(\text{\textbackslash sqrt[3]\textbraceleft 5 \textbraceright} \ = 5^{t}\). By converting the cube root to its exponential form, \(\text{5}^{1/3}\), we transform the equation into an exponential form that makes solving straightforward. Equating the exponents, since the bases are the same (5), we set \( \frac{1}{3} = t\). This is a simple example of how algebraic manipulation can be used to solve equations systematically.

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Most popular questions from this chapter

a. The populations of two countries are given for January 1,2000 , and for January 1,2010 . Write a function of the form \(P(t)=P_{0} e^{k t}\) to model each population \(P(t)\) (in millions) \(t\) years after January 1, 2000.$$ \begin{array}{|l|c|c|c|} \hline & \begin{array}{c} \text { Population } \\ \text { in 2000 } \\ \text { (millions) } \end{array} & \begin{array}{c} \text { Population } \\ \text { in 2010 } \\ \text { (millions) } \end{array} & \boldsymbol{P}(t)=\boldsymbol{P}_{0} e^{k t} \\ \hline \text { Switzerland } & 7.3 & 7.8 & \\ \hline \text { Israel } & 6.7 & 7.7 & \\ \hline \end{array}$$ b. Use the models from part (a) to predict the population on January \(1,2020,\) for each country. Round to the nearest hundred thousand. c. Israel had fewer people than Switzerland in the year 2000 , yet from the result of part (b), Israel will have more people in the year \(2020 ?\) Why? d. Use the models from part (a) to predict the year during which each population will reach 10 million if this trend continues.

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After a new product is launched the cumulative sales \(S(t)\) (in \(\$ 1000) t\) weeks after launch is given by: $$S(t)=\frac{72}{1+9 e^{-0.36 t}}$$ a. Determine the cumulative amount in sales 3 weeks after launch. Round to the nearest thousand. b. Determine the amount of time required for the cumulative sales to reach \(\$ 70,000\). c. What is the limiting value in sales?

Use the model \(A=P e^{r t} .\) The variable \(A\) represents the future value of \(P\) dollars invested at an interest rate \(r\) compounded continuously for \(t\) years. If \(\$ 10,000\) is invested in an account earning \(5.5 \%\) interest compounded continuously, determine how long it will take the money to triple. Round to the nearest year.
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