Problem 21
Question
Solve for \(x\). Give any approximate results to three significant digits. Check your answers. $$\log \left(2 x+x^{2}\right)=2$$
Step-by-Step Solution
Verified Answer
The approximate solution for \(x\) is \(x \approx 9.05\), after checking against the original equation.
1Step 1: Convert the logarithmic equation to an exponential equation
To solve the logarithmic equation \(\log (2x + x^2) = 2\), we convert it to its exponential form using the property that \(\log_b a = c \Rightarrow b^c = a\). Here, our base \(b\) is assumed to be 10 because no other base is specified. Thus, the equation \(10^2 = 2x + x^2\) is derived.
2Step 2: Simplify the exponential equation
The exponential equation \(10^2 = 2x + x^2\) simplifies to \(100 = 2x + x^2\). To solve for \(x\), we rearrange the equation into standard quadratic form \(x^2 + 2x - 100 = 0\).
3Step 3: Solve the quadratic equation
Solving the quadratic equation \(x^2 + 2x - 100 = 0\) can be done by factoring, completing the square, or using the quadratic formula. In this case, factoring doesn't yield integer solutions, so we use the quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\), where \(a = 1\), \(b = 2\), and \(c = -100\).
4Step 4: Apply the quadratic formula
Applying the quadratic formula, we have \(x = \frac{{-2 \pm \sqrt{{(2)^2 - 4(1)(-100)}}}}{{2(1)}} = \frac{{-2 \pm \sqrt{{4 + 400}}}}{{2}} = \frac{{-2 \pm \sqrt{{404}}}}{{2}}\). This gives us two potential solutions for \(x\).
5Step 5: Find the approximate solutions
By evaluating the square root, we have two approximate solutions: \(x \approx \frac{{-2 + 20.099}}{{2}}\) and \(x \approx \frac{{-2 - 20.099}}{{2}}\), which simplify to \(x \approx 9.05\) and \(x \approx -11.05\).
6Step 6: Check the solutions
We must verify that our solutions satisfy the original equation. Plugging \(x \approx 9.05\) and \(x \approx -11.05\) back into the equation, we check if \(\log(2x + x^2) = 2\). Due to the nature of logarithms, negative inputs are invalid, thus \(x \approx -11.05\) is not a valid solution. However, \(x \approx 9.05\) is valid as it produces a result close to 2 when evaluated.
Key Concepts
Exponential and Logarithmic RelationshipsQuadratic FormulaCharacteristics of Logarithms
Exponential and Logarithmic Relationships
Logarithms and exponentiation are two sides of the same coin in mathematics. They are related through an inverse operation. Understanding the relationship between these two operations is key to solving logarithmic equations.
In general, if you have a logarithmic equation of the form \(\log_b a = c\), this can be converted into an exponential equation as \(b^c = a\). This relationship allows one to switch from the logarithmic form that may seem complicated, to the exponential form which is often easier to manipulate. For example, in the exercise \(\log(2x + x^2) = 2\), we can convert this to exponentiation where the base \(b\) is assumed to be 10, resulting in \(10^2 = 2x + x^2\), which is a quadratic equation and hence, solvable using techniques for such equations.
Understanding this relationship is crucial when solving logarithmic equations because it transforms the problem into an algebraic one, where traditional techniques such as factoring, the quadratic formula, or completing the square can be applied. Always remember, when handling expressions of logarithms and exponents, the base of the logarithm matches the base of the exponent.
In general, if you have a logarithmic equation of the form \(\log_b a = c\), this can be converted into an exponential equation as \(b^c = a\). This relationship allows one to switch from the logarithmic form that may seem complicated, to the exponential form which is often easier to manipulate. For example, in the exercise \(\log(2x + x^2) = 2\), we can convert this to exponentiation where the base \(b\) is assumed to be 10, resulting in \(10^2 = 2x + x^2\), which is a quadratic equation and hence, solvable using techniques for such equations.
Understanding this relationship is crucial when solving logarithmic equations because it transforms the problem into an algebraic one, where traditional techniques such as factoring, the quadratic formula, or completing the square can be applied. Always remember, when handling expressions of logarithms and exponents, the base of the logarithm matches the base of the exponent.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It is derived from the process of completing the square, and is given by: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\).
The symbols \(\pm\) indicate that there will be two solutions, one for addition and one for subtraction. This formula is applicable in all situations involving quadratic equations, so long as the coefficients are real numbers, making it a universal solver for such equations.
Applying this to the textbook exercise, with \(a=1\), \(b=2\), and \(c=-100\), yields the potential solutions for \(x\). However, remember that each solution must be checked back in the original equation to ensure its validity. Also, be cautious of the discriminant \(b^2 - 4ac\); if it's negative, the quadratic equation has no real solutions, while a positive discriminant indicates two distinct real solutions, as was the case in the given example.
The symbols \(\pm\) indicate that there will be two solutions, one for addition and one for subtraction. This formula is applicable in all situations involving quadratic equations, so long as the coefficients are real numbers, making it a universal solver for such equations.
Applying this to the textbook exercise, with \(a=1\), \(b=2\), and \(c=-100\), yields the potential solutions for \(x\). However, remember that each solution must be checked back in the original equation to ensure its validity. Also, be cautious of the discriminant \(b^2 - 4ac\); if it's negative, the quadratic equation has no real solutions, while a positive discriminant indicates two distinct real solutions, as was the case in the given example.
Characteristics of Logarithms
Logarithms come with their own set of rules and characteristics that are fundament to their manipulation and application in solving equations. One of the critical characteristics is that a logarithm is only defined for positive real numbers. This directly impacts the validity of any solution obtained from a logarithmic equation.
For instance, even after solving the equation using algebraic methods, we are obliged to check that the solutions are within the domain of the logarithmic function. In our exercise, the solution \(x \approx -11.05\) does not satisfy the original equation \(\log(2x + x^2) = 2\) because the log of a negative number is undefined.
Moreover, other properties such as the product rule, quotient rule, and power rule for logarithms can further simplify solving logarithmic equations. Always keep these characteristics in mind as you work through logarithmic problems as they will help ensure accurate and valid results.
For instance, even after solving the equation using algebraic methods, we are obliged to check that the solutions are within the domain of the logarithmic function. In our exercise, the solution \(x \approx -11.05\) does not satisfy the original equation \(\log(2x + x^2) = 2\) because the log of a negative number is undefined.
Moreover, other properties such as the product rule, quotient rule, and power rule for logarithms can further simplify solving logarithmic equations. Always keep these characteristics in mind as you work through logarithmic problems as they will help ensure accurate and valid results.
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