Problem 18 Solve each equation. See Example... [FREE SOLUTION]

Chapter 9: Problem 18

Solve each equation. See Examples 1 through 4. $$ \log _{2}\left(x^{2}+x\right)=1 $$

Short Answer

Expert verified

The solutions are $ x = 1 $ and $ x = -2 $.

Step by step solution

Understand the Equation

We need to solve the equation $ \log_{2}(x^2 + x) = 1 $. The logarithm $ \log_{2} $ indicates that the base for the logarithm is 2.

Apply Definition of Logarithms

Recall that $ \log_{b}(a) = c $ is equivalent to $ b^c = a $. Apply this to the current problem where $ \log_{2}(x^2 + x) = 1 $. Therefore, $ 2^1 = x^2 + x $.

Simplify the Equation

Since $ 2^1 = 2 $, we substitute into the equation to get: \[ x^2 + x = 2 \] This is now a quadratic equation.

Rearrange into Standard Quadratic Form

Rearrange $ x^2 + x - 2 = 0 $ into the standard form, where the equation is $ ax^2 + bx + c = 0 $. Here, $ a = 1, b = 1$, and $ c = -2 $.

Solve the Quadratic Equation

We will solve using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute $ a = 1, b = 1, c = -2 $: \[ x = \frac{-1 \pm \sqrt{1 + 8}}{2} \] \[ x = \frac{-1 \pm \sqrt{9}}{2} \] \[ x = \frac{-1 \pm 3}{2} \]

Calculate the Solutions

Calculate the two possible solutions: 1. $ x = \frac{-1 + 3}{2} = 1 $ 2. $ x = \frac{-1 - 3}{2} = -2 $

Verify the Solutions

Check the solutions against the original equation: - For $ x = 1 $: $ \log_{2}(1+1) = \log_{2}(2) = 1 $, which is correct.- For $ x = -2 $: $ \log_{2}((-2)^2 - 2) = \log_{2}(2) = 1 $, which is also correct.

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Key Concepts

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

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree 2. It is in the form of $ ax^2 + bx + c = 0 $, where $ a $, $ b $, and $ c $ are constants. Understanding quadratic equations is crucial because they often appear in various mathematical contexts, including physics, engineering, and finance.

When solving quadratic equations, our goal is to find the value of $ x $ that makes the equation equal to zero.
These equations can have two, one, or no real solutions depending on the discriminant $ b^2 - 4ac $.

In our problem, we transformed a logarithmic equation into a quadratic equation, given by $ x^2 + x - 2 = 0 $. This step is critical because it simplifies our approach to finding $ x $ by using known algebraic methods instead of more complex logarithmic manipulations.

Logarithm Properties

Logarithms are a critical concept in algebra which help in solving equations where the variable is an exponent. Understanding the properties of logarithms can simplify complex expressions and equations:

Logarithm to a base $ b $ of a number $ a $, represented as $ \log_b{a} $, solves for $ c $ in the equation $ b^c = a $.
The inverse property of exponents and logarithms: $ b^{\log_b{a}} = a $.

In our example, applying these properties allowed us to convert the logarithmic equation $ \log_{2}(x^2 + x) = 1 $ into an exponential form $ 2^1 = x^2 + x $. By understanding such properties, we convert equations from one form to another that is more manageable for finding solutions.

Quadratic Formula

The quadratic formula is a powerful tool used for solving quadratic equations. It provides a systematic approach to finding the roots of any quadratic equation of form $ ax^2 + bx + c = 0 $. The formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

$ -b \pm \sqrt{b^2 - 4ac} $ accounts for the two potential solutions in quadratic equations.
The discriminant $ b^2 - 4ac $ determines the nature of the roots (real or complex).

In our solution, we used the quadratic formula by substituting $ a = 1 $, $ b = 1 $, and $ c = -2 $ into the formula, which yielded the roots $ x = 1 $ and $ x = -2 $. This technique ensures accuracy and completeness in solving quadratic equations.

Verifying Solutions

Verifying solutions is a critical step in solving equations to ensure correctness. Once solutions are found, they should comply with the original equation when substituted back:

Substitute each solution back into the original equation to check congruence.
Ensure no extraneous solutions exist, often a result of squaring both sides in an equation.

For this problem, we substitute both $ x = 1 $ and $ x = -2 $ back into the original logarithmic equation $ \log_{2}(x^2 + x) = 1 $. Both solutions satisfy the original equation:
- For $ x = 1 $, $ \log_{2}(2) = 1 $ matches.- For $ x = -2 $, $ \log_{2}(2) = 1 $ also holds true.
This final step confirms the solutions we found are correct and valid for the given equation.

91影视

Solve each equation. See Examples 1 through 4. $$ \log _{2}\left(x^{2}+x\right)=1 $$

Short Answer

Step by step solution

Understand the Equation

Apply Definition of Logarithms

Simplify the Equation

Rearrange into Standard Quadratic Form

Solve the Quadratic Equation

Calculate the Solutions

Verify the Solutions

Key Concepts

Quadratic Equation

Logarithm Properties

Quadratic Formula

Verifying Solutions

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