Problem 38 - Precalculus with Calculus Previews

Question

Solve the given logarithmic equation. $$ \log _{10} x^{2}+\log _{10} x^{3}+\log _{10} x^{4}-\log _{10} x^{5}=\log _{10} 16 $$

Step-by-Step Solution

Answer

The solution is \( x = 2 \).

Step 1: Simplify the Left-Hand Side

Combine the logarithms on the left-hand side of the equation using the logarithm property that combines the sum or difference of logs: \( \log_b a + \log_b c = \log_b (a \cdot c) \) and \( \log_b a - \log_b c = \log_b \left( \frac{a}{c} \right) \). This gives:\[ \log_{10} (x^2 \cdot x^3 \cdot x^4) - \log_{10} (x^5) = \log_{10} 16 \]

Step 2: Further Simplify Using Exponents

Simplify the expression \( (x^2 \cdot x^3 \cdot x^4) \) using exponent rules: \( x^a \cdot x^b = x^{a+b} \). This gives \( x^{2+3+4} = x^9 \). So now, the left-hand side becomes: \[ \log_{10} \left( \frac{x^9}{x^5} \right) = \log_{10} 16 \]

Step 3: Simplify the Fraction in the Logarithm

Simplify the fractional exponent using the rule: \( \frac{x^a}{x^b} = x^{a-b} \). Applying this, we get: \( \frac{x^9}{x^5} = x^{9-5} = x^4 \). Thus, the equation is now: \[ \log_{10} x^4 = \log_{10} 16 \]

Step 4: Solve for the Variable

Since the logarithms are equal, the arguments must be equal as well. Therefore, set the arguments equal to each other: \( x^4 = 16 \).

Step 5: Solve the Exponential Equation

Recognize that \( 16 \) can be rewritten as \( 2^4 \). So, the equation \( x^4 = 2^4 \) implies \( x = 2 \). Check for additional solutions based on the domain of logarithmic functions. Since logarithms are defined for positive numbers, \( x \) must be positive. Hence, solution is complete.

Key Concepts

Properties of Logarithms

Logarithms have several properties that make solving logarithmic equations much easier. One key property is the product rule, which states that the logarithm of a product is equal to the sum of the logarithms of its factors:

• \( \log_b (a \cdot c) = \log_b a + \log_b c \)

This allows us to combine multiple logarithmic terms into a single term. Similarly, the quotient rule helps us deal with differences:

• \( \log_b \left( \frac{a}{c} \right) = \log_b a - \log_b c \)

These properties are extremely useful when simplifying complex logarithmic expressions, like in our example where multiple logs were combined and simplified into a single logarithm.

Exponent Rules

Understanding the rules of exponents is critical when working with logarithms, as logarithmic functions are the inverse of exponential functions. The fundamental rule for multiplying powers with the same base is:

• \( x^a \cdot x^b = x^{a+b} \)

This allows us to combine the terms effectively. For example, in the original problem, terms \( x^2 \), \( x^3 \), and \( x^4 \) were combined into \( x^9 \) using this rule, simplifying the expression considerably.
Similarly, when dividing powers, we use:

• \( \frac{x^a}{x^b} = x^{a-b} \)

This was applied to transform \( \frac{x^9}{x^5} \) into \( x^4 \). Knowing these rules provides a smooth pathway to simplify and solve exponential and logarithmic equations.

Solving Exponential Equations

An exponential equation is one in which variables appear as exponents. Solving these equations often involves matching the bases or using logarithms to bring exponents down. In the exercise, once the expression \( \log_{10} x^4 = \log_{10} 16 \) was simplified, the problem boiled down to matching the exponents since the bases were the same.
Recognizing that both sides could be written in powers of 2 helped find the solution:

• \( x^4 = 2^4 \) implies \( x = 2 \)

We looked at the domain considerations for the solution to ensure all values were valid within the context of the logarithm function, which only takes positive arguments. Thus, checking and confirming the solution's validity is a crucial final step.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. They have their domain primarily in the positive real numbers, which means the results from solving logarithmic equations must also satisfy this condition.
In our solved example, the function was used to transform a complex combination of powers into a simpler equation through the properties of logarithms and exponents.
Understanding the behavior and restrictions of logarithmic functions is crucial. For example, since no logarithm of a negative number or zero exists in real numbers, we concluded by verifying that our solution \( x = 2 \) is a positive real number, ensuring the integrity of the solution within the domain of logarithmic functions.