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Chapter 7: Problem 16

In Exercises \(13-16,\) what happens to \(y\) when \(x\) is doubled? Here \(k\) is a positive constant. $$ \frac{y}{x^{4}}=k $$

Short Answer

Expert verified
Answer: When \(x\) is doubled, the value of \(y\) becomes 16 times greater.

Step by step solution

01

Solving for y

First, let's solve the equation for \(y\) in terms of \(x\) and the constant \(k\): $$ \frac{y}{x^{4}} = k $$ To isolate \(y\), multiply both sides by \(x^4\): $$ y = kx^4 $$ Now we have an equation relating \(y\) and \(x\).
02

Doubling x

Next, let's see what happens when we double the value of \(x\). We'll replace \(x\) with \((2x)\) in the equation and see how it affects \(y\): $$ y = k(2x)^4 $$
03

Simplify the equation

Now, let's simplify the equation to find the relationship between \(y\) and the new value of \(x\): $$ y = k(16x^4) $$ Observe that the right-hand side of the equation has changed from \(kx^4\) to \(16kx^4\) when we doubled the value of \(x\).
04

Comparing y-values

Contemplate how this affects the value of \(y\). When \(x\) is doubled, the right-hand side of the equation becomes \(16\) times greater: $$ y = 16kx^4 $$
05

Conclusion

Thus, when \(x\) is doubled, the value of \(y\) becomes 16 times greater.

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Key Concepts

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

Exponentiation
Exponentiation is like supercharging a number. It allows you to multiply it by itself a certain number of times. This is commonly represented as a base raised to an exponent, written as \( x^n \). Here, \( x \) is the base and \( n \) tells you how many times to multiply it. In the exercise, we used exponentiation with \( x^4 \), which means multiplying \( x \) by itself four times: \( x \times x \times x \times x \). When you double \( x \) and then raise it to the fourth power as \( (2x)^4 \), you multiply 2 by itself four times, resulting in 16. Hence, the term \( x^4 \) in the expression for \( y \) becomes \( 16x^4 \).
  • Exponentiation helps you perform multiple multiplications quickly.
  • Doubling a base before raising it to a power can greatly increase the result.
  • It's a core concept helping to understand growth, scaling, and other phenomena.
Direct Variation
Direct variation describes a relationship between two variables where one is a constant multiple of the other. In simpler terms, when one variable increases, the other increases at a constant rate. In the form of the equation \( y = kx^n \), \( k \) is a constant, and the change in \( x \) directly influences \( y \). For our exercise, \( y \) varies directly with \( x^4 \), which means as \( x \) changes, \( y \) will change in a predictable way dependent on \( k \).
  • Direct variation allows prediction and control of outcomes by understanding one variable influences another directly.
  • Improvements in technology, science, and predictions often rely on identifying and leveraging direct variations.
  • It's fundamental in both linear and nonlinear relationships.
Equation Solving
Equation solving is like figuring out a puzzle. It involves finding the value of unknowns that make the equation true. We start by isolating the unknown variable, which in our exercise was \( y \). By solving the initial equation \( \frac{y}{x^4} = k \), we multiplied both sides by \( x^4 \) to isolate \( y \), giving us \( y = kx^4 \). When we double \( x \), we substitute this new value into our equation, then simplify through exponentiation: \( y = k(2x)^4 = 16kx^4 \). This shows that doubling changes the equation's entire balance and relationship.
  • Equation solving makes unknowns known by revealing relationships between variables.
  • This approach is crucial in science, business, and everyday problem solving.
  • Understanding the step-by-step solution process is vital for mastering algebra.

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Most popular questions from this chapter

\(a\) and \(b\) are positive numbers and \(a>b\). Which is larger, \(f(a)\) or \(f(b)\) ? $$ f(t)=-3 t^{2}\left(-2 t^{3}\right)^{5} $$

If \(z\) is proportional to a power of \(y\) and \(y\) is proportional to a power of \(x\), is \(z\) proportional to a power of \(x\) ?

Without solving them, say whether the equations in Exercises \(27-42\) have (i) One positive solution (ii) One negative solution (iii) One solution at \(x=0\) (iv) Two solutions (v) \(\quad\) Three solutions (vi) No solution Give a reason for your answer. $$ x^{1 / 3}=-2 $$

A certificate of deposit is worth \(P(1+r)^{t}\) dollars after \(t\) years, where \(r\) is the annual interest rate expressed as a decimal, and \(P\) is the amount initially deposited. State which investment will be worth more. Investment \(A,\) in which \(P=\$ 10,000, r=3 \%,\) and \(t=30\) years or investment \(B,\) in which \(P=\$ 15,000\), \(r=4 \%,\) and \(t=15\) years.

In Exercises \(1-4\), write a formula for \(y\) in terms of \(x\) if \(y\) satisfies the given conditions. Proportional to the \(4^{\text {th }}\) power of \(x,\) and \(y=10.125\) when \(x=3\).
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