/*! 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 87 Solve each equation. $$(4 y-5)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 87

Solve each equation. $$(4 y-5)^{7}=0$$

Short Answer

Expert verified
y = \frac{5}{4}

Step by step solution

01

Understand the Problem

We are given the equation \( (4y - 5)^7 = 0 \). This equation is asking us to find the value of \( y \) that makes the entire expression equal to zero.
02

Simplify the Equation

Since the exponent 7 is applied to the entire expression, for it to equal zero, the base (inside the parentheses) must be zero. Thus, we need to solve the equation \( 4y - 5 = 0 \).
03

Solve for y

Rearrange the equation \( 4y - 5 = 0 \) to solve for \( y \). Add 5 to both sides: \( 4y = 5 \), and then divide both sides by 4: \( y = \frac{5}{4} \).

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.

Zero Exponent Rule
The zero exponent rule states that any base with an exponent of zero is equal to one. Mathematically, this is written as: \(a^0 = 1\), where \(a\) is any non-zero number.
This rule is crucial when simplifying equations with exponents. In our exercise, the equation turned out to have an exponent, but because the result equals zero, the inside of the parentheses must be zero, not the entire expression raised to that power.
Remember, in multiplication and powers, if any factor is zero, the entire product is zero.
  • Exponential rules simplify complex equations.
  • They help identify critical base components.
  • Zero exponent rule clarifies the necessity of non-zero bases.
Linear Equations
Linear equations form the basis of algebra. They describe relationships where one variable is directly proportional to another, forming a straight line when graphed.
In a linear equation, unknowns appear to the first power (no exponents involved). A standard linear equation is in the form of \(ax + b = c\).
Let's break down our exercise: In the step where \(4y - 5 = 0\), we see a clear example of a linear equation. To solve,
  • Isolate the variable (\(y\) in this case).
  • Perform inverse operations.
  • Balance both sides of the equation.
By adding 5 to both sides and then dividing by 4, we isolate \(y\).
This gives us \(y = \frac{5}{4}\), proving the simplicity and power of linear equations.
Algebraic Manipulation
Algebraic manipulation involves rearranging equations to isolate and solve for variables.
It uses various operations: addition, subtraction, multiplication, and division. In our exercise, manipulating \(4y - 5 = 0\) included steps:
  • Adding 5 to both sides: \(4y = 5\).
  • Dividing both sides by 4: \(y = \frac{5}{4}\).
This systematic approach not only finds the solution but also maintains the equation's balance.
Some tips for algebraic manipulation include:
  • Performing the same operation on both sides.
  • Keeping track of positive and negative signs.
  • Simplifying fractions and combining like terms.
With these techniques, solving even complex problems can become methodical and straightforward.

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 each problem by writing an equation and solving it. Find the exact answer and simplify it using the rules for radicals. Average annual return. $$r=\left(\frac{S}{P}\right)^{1 / n}-1$$ was used to find the average annual return for an investment. a) Write \(S\) as a function of \(r, P,\) and \(n\) b) Write \(P\) as a function of \(r, S,\) and \(n\)

Solve each equation. $$w^{-2 / 3}=4$$

Solve each equation. $$t^{-1 / 2}=9$$

Simplify. $$\frac{3 \sqrt{k}}{\sqrt{k}+\sqrt{7}}$$

Simplify. $$\frac{\sqrt{h k}}{\sqrt{h}+3 \sqrt{k}}$$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision 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.