Problem 46 - Algebra and Trigonometry

Question

Solve the logarithmic equation for \(x\) $$ \log _{5} x+\log _{5}(x+1)=\log _{5} 20 $$

Step-by-Step Solution

Answer

The solution is \(x = 4\).

Step 1: Use the Product Rule for Logarithms

The equation given is \( \log _{5} x + \log _{5}(x+1) = \log _{5} 20 \). Use the product rule for logarithms, which states \( \log_b a + \log_b c = \log_b (a \cdot c) \). Apply it to the left side: \( \log_5 (x \cdot (x+1)) = \log_5 20 \).

Step 2: Drop the Logarithms

Since the logarithms are to the same base on both sides, drop the logarithms: \( x(x+1) = 20 \). This simplifies to the equation \( x^2 + x = 20 \).

Step 3: Rearrange the Equation

Rearrange the equation into standard quadratic form: \( x^2 + x - 20 = 0 \).

Step 4: Factor the Quadratic Equation

Look for two numbers that multiply to \(-20\) and add up to \(1\). The numbers are \(5\) and \(-4\). The factored form of the equation is \((x+5)(x-4) = 0\).

Step 5: Solve for x

Set each factor equal to zero and solve for \(x\): \(x+5=0\) yields \(x = -5\). \(x-4=0\) yields \(x = 4\).

Step 6: Verify Solutions in the Logarithmic Context

Check the potential solutions \(x = -5\) and \(x = 4\) back in the original logarithmic equation. The logarithm of a negative number is undefined, so \(x = -5\) is not valid. \(x = 4\) is valid because \(\log_5 4\) and \(\log_5 5\) are defined.

Key Concepts

Product Rule in Logarithmic Equations

Logarithmic equations often utilize rules of logarithms to simplify and solve them. One fundamental tool is the Product Rule. This rule states that the sum of two logarithms with the same base can be combined into a single logarithm. It is expressed as: \( \log_b a + \log_b c = \log_b (a \cdot c) \). This is especially handy in our equation:

• Start with: \( \log_5 x + \log_5(x+1) \).
• Use the product rule: \( \log_5 (x \cdot (x+1)) \).
• Simplified to: \( \log_5 20 \).

Now the equation is a single logarithm on both sides. This allows for further simplification by removing the logs.

Understanding Quadratic Equations

Once the logarithms are removed, the equation transforms into a quadratic equation. Quadratic equations are polynomial equations of degree two, typically written in the standard form: \( ax^2 + bx + c = 0 \).

• In our case, we derive \( x^2 + x = 20 \) after eliminating the logs.
• Rearrange to: \( x^2 + x - 20 = 0 \) to fit the standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = 1 \), and \( c = -20 \).

Quadratic equations are pivotal as they often arise in various branches of mathematics and have specific methods of solving, like factoring, which is our next step.

Factoring Quadratic Equations

Factoring is a method of solving quadratic equations that involves writing the equation as a product of its factors set to zero. In our example, we need to find two numbers that multiply to \(-20\) and add up to \(1\).

• These numbers are: \(5\) and \(-4\).
• This means the equation can be factored as: \((x+5)(x-4) = 0\).

Through factoring, the quadratic equation is broken down into its simplest form. Each factor is then set to zero, leading us to solve for potential values of \(x\). This efficient method is widely used as it simplifies complex equations into manageable parts.

Verification of Solutions

The final step in solving equations, especially logarithmic ones, is to verify the solutions. This ensures that the solutions are applicable in the original equation and are mathematically valid.

• The solutions from \((x+5)(x-4) = 0\) are \(x = -5\) and \(x = 4\).
• Test these values in the context of the original logarithmic equation.
• For \(x = -5\), since logarithms of negative numbers are undefined, it is discarded.
• For \(x = 4\), substitute back to see if it satisfies \( \log_5 x + \log_5(x+1) = \log_5 20 \).

Only \(x = 4\) meets the criteria, confirming it as a valid solution. This verification step is crucial to ensure that the found solutions are not only mathematically correct but also make logical sense in the context of the problem.