Problem 38
Question
Solve each equation with rational exponents. Check all proposed solutions. $$(x+5)^{\frac{2}{3}}-4$$
Step-by-Step Solution
Verified Answer
The original equation has no solution.
1Step 1: Isolate the term with the rational exponent
First, we add 4 to both sides of the equation to get: \((x+5)^{\frac{2}{3}}=4\).
2Step 2: Remove the rational exponent
We can get rid of the rational exponent by raising both sides of the equation to the power of \( \frac{3}{2} \). This gives us: \((x+5)^{\frac{2}{3}*\frac{3}{2}}=(4)^{\frac{3}{2}}\).
3Step 3: Simplify the equation
Simplify both sides of the equation: \((x+5)^1 = 8\).
4Step 4: Solve for x
Finally, subtract 5 from both sides of the equation to solve for x. This gives \(x=8-5 = 3\).
5Step 5: Check the proposed solution
Plug \(x=3\) back into the original equation to ensure it is a valid solution. \((3+5)^{\frac{2}{3}}-4 = 2-4 = -2\), which is not equal to zero. Thus, the original equation has no solution.
Key Concepts
Rational ExponentsIsolating the VariableExponentiationChecking Solutions in Algebra
Rational Exponents
Rational exponents are a way to express roots as powers. For instance, the expression \(a^{\frac{m}{n}}\) represents the nth root of a raised to the mth power. Understanding this concept is crucial when solving equations that involve roots, like in our exercise where we have the term \(x+5)^{\frac{2}{3}}\). This notation simplifies expressing and manipulating roots within algebraic operations.
To manipulate expressions with rational exponents, we can use the laws of exponents to simplify and solve equations. A core principle is that raising a power to another power means you multiply exponents, such as \(\left(a^m\right)^n = a^{m\times n}\). This principle was applied in our exercise when we raised both sides of the equation to the power of \(\frac{3}{2}\) to eliminate the rational exponent.
To manipulate expressions with rational exponents, we can use the laws of exponents to simplify and solve equations. A core principle is that raising a power to another power means you multiply exponents, such as \(\left(a^m\right)^n = a^{m\times n}\). This principle was applied in our exercise when we raised both sides of the equation to the power of \(\frac{3}{2}\) to eliminate the rational exponent.
Isolating the Variable
Isolating the variable is a fundamental skill in algebra. It means rearranging the equation so that the variable of interest is alone on one side of the equation. The goal is to discover the value of this variable. In our exercise, we started with the equation \( (x+5)^{\frac{2}{3}}-4 = 0\) and isolated the term with the rational exponent by adding 4 to both sides, obtaining \( (x+5)^{\frac{2}{3}} = 4\).
To correctly isolate the variable, you must perform the same action on both sides of the equation to keep it balanced. It's like a scale: what you do to one side, you must do to the other. After isolation, you can proceed to solve for the variable using various algebraic techniques.
To correctly isolate the variable, you must perform the same action on both sides of the equation to keep it balanced. It's like a scale: what you do to one side, you must do to the other. After isolation, you can proceed to solve for the variable using various algebraic techniques.
Exponentiation
Exponentiation is the process of raising a number to a given power. It is a fundamental operation in algebra that is used to describe repeated multiplication. When a number with a rational exponent is involved, exponentiation can help us get rid of the exponent and simplify the equation for easier solving.
In the case of our exercise, we had to deal with the expression \(4^\frac{3}{2}\). This means the square root of 4, which is 2, raised to the power of 3, resulting in 8. When we raised both sides of our equation to the power of \(\frac{3}{2}\), the left side's exponent became 1 (since \(\frac{2}{3} \times \frac{3}{2} = 1\)), effectively removing the rational exponent and giving us a linear equation that is easier to solve.
In the case of our exercise, we had to deal with the expression \(4^\frac{3}{2}\). This means the square root of 4, which is 2, raised to the power of 3, resulting in 8. When we raised both sides of our equation to the power of \(\frac{3}{2}\), the left side's exponent became 1 (since \(\frac{2}{3} \times \frac{3}{2} = 1\)), effectively removing the rational exponent and giving us a linear equation that is easier to solve.
Checking Solutions in Algebra
After finding a proposed solution to an algebraic equation, it is essential to check that the solution is valid within the context of the original equation. This simply means substituting the solution back into the original equation to see if a true statement is produced.
In our exercise, the solution of \(x = 3\) was obtained after solving the simplified equation. However, when we substituted \(x = 3\) back into the original equation, we found that the result did not satisfy the equation because it did not equal zero. Thus, despite the solution appearing correct in the isolated form, the original equation tells us it is not valid. This crucial step ensures any extraneous solutions introduced through the process of solving – particularly when dealing with rational exponents and roots – are identified and discarded.
In our exercise, the solution of \(x = 3\) was obtained after solving the simplified equation. However, when we substituted \(x = 3\) back into the original equation, we found that the result did not satisfy the equation because it did not equal zero. Thus, despite the solution appearing correct in the isolated form, the original equation tells us it is not valid. This crucial step ensures any extraneous solutions introduced through the process of solving – particularly when dealing with rational exponents and roots – are identified and discarded.
Other exercises in this chapter
Problem 38
Exercises \(37-38\) involve markup, the amount added to the dealer's cost of an item to arrive at the selling price of that item. The selling price of a scienti
View solution Problem 38
Perform the indicated operations and write the result in standard form. $$ \frac{-12+\sqrt{-28}}{32} $$
View solution Problem 39
Use interval notation to express solution sets and graph each solution set on a number line. Solve linear inequality. \(1-(x+3) \geq 4-2 x\)
View solution Problem 39
contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These a
View solution