Problem 87 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \frac{\log (5 x+6)}{2}=\log x $$

Step-by-Step Solution

Answer

The solution is \(x = 6\).

Step 1: Eliminate the Fraction

Multiply both sides of the equation by 2 to eliminate the fraction:\[\log(5x + 6) = 2 \log x\] This simplifies the equation and gets rid of the fraction.

Step 2: Apply Logarithm Rules

Use the power rule of logarithms on the right side:\[\log(5x + 6) = \log(x^2)\] This uses the rule \(b\log(a) = \log(a^b)\).

Step 3: Set the Logarithms Equal

Since the logarithms are equal, set the contents equal as well:\[5x + 6 = x^2\] This follows from the property that if \(\log(a) = \log(b)\), then \(a = b\).

Step 4: Rearrange into a Quadratic Equation

Rearrange the equation to form a standard quadratic equation:\[x^2 - 5x - 6 = 0\] Move all terms to one side.

Step 5: Solve the Quadratic Equation

Factor the quadratic equation:\[(x - 6)(x + 1) = 0\] Therefore, the solutions are:\[x = 6 \quad \text{and} \quad x = -1\]

Step 6: Verify Solutions

Ensure both solutions are valid by substituting back into the original equation. Substituting \(x = -1\) gives a negative argument for the logarithm, which is not possible.- Check \(x = 6\): - \(\log(5(6) + 6) / 2 = \log(6)\) - Simplifies to \(\log 36 / 2 = \log 6\) - \(\log 6 = \log 6\) is true, so \(x = 6\) is valid.

Key Concepts

Quadratic Equations

Quadratic equations are polynomials of the form \(ax^2 + bx + c = 0\) where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations have two solutions that can be found using methods such as factoring, completing the square, or the quadratic formula. In our equation \(x^2 - 5x - 6 = 0\), factoring is a straightforward method:

• Identify two numbers that multiply to \(-6\) and add to \(-5\). These numbers are \(-6\) and \(1\).
• Write the equation as \((x - 6)(x + 1) = 0\).
• Set each factor equal to zero: \(x - 6 = 0\) and \(x + 1 = 0\).
• Solve these equations to find \(x = 6\) and \(x = -1\).

Not all quadratic equations will factor easily. In such cases, other methods may be used. Always verify your solutions in the context of the problem, such as checking for restrictions like negative values under a logarithm.

Logarithm Properties

Logarithms have several important properties that help simplify expressions. Understanding these properties is key to solving equations involving logs:

• Product property: \(\log(a) + \log(b) = \log(ab)\)
• Quotient property: \(\log(a) - \log(b) = \log\left(\frac{a}{b}\right)\)
• Power property: \(b\log(a) = \log(a^b)\)

In the exercise, we applied the power property as follows: \(2 \log(x) = \log(x^2)\). This allowed us to equate two logarithmic expressions and solve for \(x\). Note that logarithms are undefined for non-positive arguments, hence only positive values should be considered.

Exact Solutions

An exact solution is one expressed in terms of integers, fractions, or known constants like \(\pi\). When solving equations, obtaining exact solutions is often more desirable before approximating if necessary. For example, the exact solution to our problem was \(x = 6\). Approximations are less precise and are used when necessary for practical applications.In this exercise, the only valid exact solution was \(x = 6\), as \(x = -1\) resulted in an invalid logarithmic argument, showcasing the importance of understanding problem constraints. Exact solutions provide the precise mathematical solution that perfectly satisfies the equation without rounding.