Problem 91 - Intermediate Algebra

Question

For Problems \(81-97\), solve each of the equations. $$ \log _{8}(x+7)+\log _{8} x=1 $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Use the logarithmic addition rule

Start by applying the product rule for logarithms which states that \( \log_b(a) + \log_b(b) = \log_b(a \cdot b) \). So we have: \( \log_8((x+7) \cdot x) = 1 \).

Step 2: Simplify the expression

Combine the terms inside the logarithm: \( \log_8(x^2 + 7x) = 1 \).

Step 3: Convert the logarithmic equation to an exponential equation

Use the definition of a logarithm: if \( \log_b(y) = c \), then \( y = b^c \). Apply this to our equation: \( x^2 + 7x = 8^1 \), which simplifies to \( x^2 + 7x = 8 \).

Step 4: Form a quadratic equation

Bring all the terms to one side of the equation: \( x^2 + 7x - 8 = 0 \).

Step 5: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 7, c = -8 \). Calculate the discriminant: \( b^2 - 4ac = 49 + 32 = 81 \). Substitute into the formula to find the roots: \( x = \frac{-7 \pm 9}{2} \).

Step 6: Find the roots

Calculate the two potential solutions: \( x = \frac{2}{2} = 1 \) and \( x = \frac{-16}{2} = -8 \).

Step 7: Verify the solutions

Check each root to ensure it satisfies the original equation. Notice that \( x = -8 \) is not valid since you cannot take the log of a negative or zero. Therefore, \( x = 1 \) is the only valid solution.

Key Concepts

Logarithmic Rules

When we deal with logarithmic equations, it's essential to understand a few fundamental rules that simplify calculations. Logarithms are basically the inverse operations of exponentials, and they follow specific rules that help us solve equations efficiently.One commonly used rule is the **product rule**, which states that the sum of two logarithms with the same base is equivalent to the logarithm of the product of their arguments:

• \[\log_b(a) + \log_b(c) = \log_b(a \, c)\]

In the original exercise, this rule was used to combine \(\log_8(x+7)\) and \(\log_8x\) into \(\log_8((x+7) \cdot x)\).Logarithms also have a **power rule** and a **quotient rule** which are used in other types of logarithmic calculations:

• **Power Rule**: \( \log_b(a^n) = n \cdot \log_b(a) \)
• **Quotient Rule**: \( \log_b( \frac{a}{c}) = \log_b(a) - \log_b(c) \)

Understanding these rules makes it much easier to manipulate and solve logarithmic equations by simplifying the expressions before solving.

Quadratic Equations

Quadratic equations are central to many algebra problems and often appear in mathematical calculations involving real-world scenarios. A quadratic equation generally takes the form:

• \[ ax^2 + bx + c = 0 \]

In our exercise, after converting the logarithmic equation, we ended up with a quadratic equation: \(x^2 + 7x - 8 = 0\).Quadratic equations can be solved by various methods such as factoring, using the quadratic formula, or completing the square. The quadratic formula is very useful, especially when a quadratic expression doesn't factor neatly:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In our problem, substituting the values \(a = 1\), \(b = 7\), and \(c = -8\) into the formula gives us the roots of the equation. After calculating, we get potential solutions, but only one of them, \(x = 1\), is valid after verification.

Exponential Equations

To solve our original logarithmic equation, we had to convert it to an exponential form. This is done using the relationship between logarithms and exponentials, which is fundamental in solving such equations.When we have a logarithmic equation like \(\log_b(y) = 1\), its equivalent exponential form is \(y = b^1\). By rewriting the equation from logarithmic to exponential form, we can often solve for the variable using standard algebraic methods.In our exercise, we converted \(\log_8(x^2 + 7x) = 1\) into \(x^2 + 7x = 8\) which then allowed us to solve it as a quadratic equation.Exponential equations can either increase or decrease very quickly, which is why understanding how to manipulate them, often involving logarithms, is crucial. They come into play in topics ranging from interest calculations to scientific growth and decay models. In many cases, logarithms offer a way to solve exponential equations by converting them into simpler forms like linear or quadratic equations.