/*! 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 65 Solve. Where appropriate, includ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve. Where appropriate, include approximations to three decimal places. If no solution exists, state this. $$ \log _{2}(x-2)+\log _{2} x=3 $$

Short Answer

Expert verified
x = 4.

Step by step solution

01

Use the properties of logarithms

Using the property of logarithms that \(\log_b m + \log_b n = \log_b (mn)\), we can combine the two logarithms on the left side: \[\log_{2} (x-2) + \log_{2} x = \log_{2} ((x-2)x).\]
02

Simplify the logarithmic equation

The equation now is: \[\log_{2} ((x-2)x) = 3.\] Simplify inside the logarithm: \[x(x-2) = x^2 - 2x.\]
03

Remove the logarithm

To isolate \[x^2 - 2x, \] rewrite the equation in exponential form: \[2^3 = x^2 - 2x.\] This simplifies to: \[8 = x^2 - 2x.\]
04

Form a quadratic equation

Rewrite the equation to form a standard quadratic equation: \[x^2 - 2x - 8 = 0.\]
05

Solve the quadratic equation

We can solve \[x^2 - 2x - 8 = 0\] either by factoring, using the quadratic formula, or completing the square. Factoring gives us: \[(x-4)(x+2) = 0.\] So, \[x - 4 = 0 \] or \[x + 2 = 0,\] providing solutions: \[x = 4\] or \[x = -2.\]
06

Check the solutions

Since \[x = -2\] would result in taking a logarithm of a negative number or zero, which is undefined, only \[x = 4\] is a valid solution.

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 91Ó°ÊÓ!

Key Concepts

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

Properties of Logarithms
Logarithms have several properties that are useful for solving equations. One key property is the Product Rule, which states \( \log_b(m) + \log_b(n) = \log_b(mn) \). This means that if you have the logarithms of two numbers added together, you can combine them into one logarithm by multiplying the numbers inside. This property was used in the exercise to combine \( \log_{2}(x-2) + \log_{2} x \) into \( \log_{2} ((x-2)x) \). Another important property is that the logarithm of a product is still a product. Always make sure to combine logs where possible, as it reduces the complexity of the equation. This step is crucial as it sets up the equation to be simplified further, helping you get closer to the solution. Remember: the goal is to apply these properties to present the equation in a simpler form.
Solving Quadratic Equations
Quadratic equations have the general form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. To solve a quadratic equation, you can use different methods such as factoring, the quadratic formula, or completing the square. In the given exercise, the equation \( 8 = x^2 - 2x \) was rearranged into a standard quadratic form: \( x^2 - 2x - 8 = 0 \). Factoring was chosen as the solution method, providing factors \( (x-4)(x+2) = 0 \). To find the solutions, you set each factor to zero: \( x-4 = 0 \) and \( x+2 = 0 \), resulting in \( x = 4 \) and \( x = -2 \). It's essential to verify the solutions afterward, as with logarithms, some solutions might be extraneous.
Logarithmic Functions
A logarithmic function is the inverse of an exponential function. The general form of a logarithmic function is \( y = \log_b(x) \), where \( b \) is the base and \( y \) is the result. Logarithmic functions are useful because they can turn multiplicative relationships into additive relationships, making complex problems easier to manage. In our exercise, understanding the behavior of the logarithmic function \( \log_{2}(x-2) + \log_{2} x = 3 \) was crucial. By converting the added logarithms into a single log expression and changing this equation into exponential form, the problem became much easier to solve. This transformation highlights one of the greatest strengths of logarithms: simplifying multiplicative processes.
Exponential and Logarithmic Forms
The connection between exponential and logarithmic forms is essential in solving logarithmic equations. Any logarithmic equation \( \log_b(y) = x \) can be rewritten in its exponential form as \( b^x = y \). For instance, in the exercise, \( \log_{2}((x-2)x) = 3 \) was converted into \( 2^3 = (x-2)x \), simplifying it to \( 8 = x^2 - 2x \). This step was crucial, transforming a complicated logarithmic equation into a straightforward quadratic equation. Understanding this relationship helps in maneuvering between logarithmic and exponential equations, enabling more flexible problem-solving strategies and simplifying equations step-by-step.

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

Find each of the following, given that $$ f(x)=\frac{1}{x+2} \quad \text { and } \quad g(x)=5 x-8 $$ $$ f(-1) $$

The decay rate of krypton- 85 is \(6.3 \%\) per year. What is its half-life?

As of May \(2016,\) Giancarlo Stanton of the Miami Marlins had the largest contract in sports history. As part of the 13 -year 325 million dollars deal, he will receive 32 million dollars in \(2023 .\) How much money would need to be invested in 2015 at \(4 \%\) interest, compounded continuously, in order to have 32 million dollars for Stanton in \(2023 ?\) (This is much like determining what 32 million dollars in 2023 is worth in 2015 dollars.) Data: Forbes.com

Determine whether or not the given pairs of functions are inverses of each other. \(f(x)=0.8 x^{1 / 2}+5.23\) \(g(x)=1.25\left(x^{2}-5.23\right), x \geq 0\)

The concentration of acetaminophen in the body decreases exponentially after a dosage is given. In one clinical study, adult subjects averaged 11 micrograms \(/\) milliliter \((\mathrm{mcg} / \mathrm{mL})\) of the drug in their blood plasma 1 hr after a 1000 -mg dosage and 2 micrograms / milliliter 6 hr after dosage. Data: tylenolprofessional.com; Mark Knopp, M.D. a) Find the value \(k,\) and write an equation for an exponential function that can be used to predict the concentration of acetaminophen, in micrograms / milliliter, t hours after a 1000 -mg dosage. b) Estimate the concentration of acetaminophen 3 hr after a 1000 -mg dosage. c) To relieve a fever, the concentration of acetaminophen should go no lower than \(4 \mathrm{mcg} / \mathrm{mL}\). After how many hours will a 1000 -mg dosage drop to that level? d) Find the half-life of acetaminophen.

See all solutions

Recommended explanations on Math Textbooks

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.