Problem 46 - Algebra A Combined Function

Question

Solve each equation. Give an exact solution and a four-decimal-place approximation. $$ \log (3 x-2)=-0.8 $$

Step-by-Step Solution

Answer

Exact solution: \( x = \frac{21.585}{30} \); Approximate solution: \( x \approx 0.7195 \).

Step 1: Understand the Equation

We start with the equation \( \log(3x - 2) = -0.8 \). The logarithm is in base 10 here (common logarithm). Our goal is to solve for \( x \).

Step 2: Convert Logarithmic Equation to Exponential Form

Recall that if \( \log_b(A) = C \), then \( A = b^C \). Applying this here, we convert the equation \( \log(3x - 2) = -0.8 \) to exponential form, becoming \( 3x - 2 = 10^{-0.8} \).

Step 3: Calculate the Power of 10

Compute \( 10^{-0.8} \) using a calculator, which gives an approximate value of 0.1585 (rounded to four decimal places).

Step 4: Solve for x

Now solve the equation \( 3x - 2 = 0.1585 \). Add 2 to both sides: \( 3x = 2.1585 \). Then, divide both sides by 3 to find \( x \): \( x = \frac{2.1585}{3} \).

Step 5: Calculate the Exact and Approximate Solution

Calculate \( x = \frac{2.1585}{3} \), which exactly is \( x = \frac{21.585}{30} \) and approximately is \( x \approx 0.7195 \) (rounded to four decimal places).

Key Concepts

Exponential Form

In solving logarithmic equations, converting them to exponential form can greatly simplify the process. The rule we use is that if \( \log_b(A) = C \), this simply means that \( A = b^C \). Here, \( b \) is the base of the logarithm. For the given problem, we have \( \log(3x - 2) = -0.8 \) in base 10, which upon conversion gives us the exponential equation \( 3x - 2 = 10^{-0.8} \). By converting to exponential form, we can handle the equation like a regular algebraic equation, making it easier to solve for the unknown variable. This step is crucial, as it transforms a complex logarithmic form into a more straightforward equation that is easier to solve.

Common Logarithm

Understanding common logarithms is important for solving exercises like this. Common logarithms are base 10 logarithms and are often simply denoted as \( \log(x) \) instead of \( \log_{10}(x) \). This is because base 10 is the most frequently used in everyday calculations, especially in scientific and engineering contexts.The current problem \( \log(3x - 2) = -0.8 \) uses a common logarithm. Recognizing that "log" without a base usually implies base 10 can prevent confusion when manipulating the equation. The use of base 10 makes it easier to use standard calculators, as most are designed to handle this logarithmic base efficiently.

Approximation

In mathematics, sometimes finding an exact solution involves complicated fractions or irrational numbers, and an approximate solution becomes useful for practical calculations. After converting the logarithmic equation to exponential form and solving, we reached the expression \( 3x - 2 = 10^{-0.8} \).When evaluated, \( 10^{-0.8} \) approximately equals 0.1585, providing a numerical solution that can be more practical for everyday use. While exact solutions have their place, approximations are often necessary to communicate results that are easily interpretable and applicable.Be sure to round correctly to the specified number of decimal places, in this case, four, for precision and accuracy in your work.

Base 10

Base 10 is a fundamental concept in logarithms and is crucial for understanding common logarithms. A logarithm with base 10 means that the power of 10 needed to yield a given number is represented. Since base 10 is involved, the whole logarithmic and exponential concepts hinge on powers of 10. For instance, \( 10^1 = 10 \), \( 10^0 = 1 \), and \( 10^{-0.8} \) which we calculated earlier.Base 10 logarithms are particularly intuitive because our numerical system is base 10. This makes calculations with base 10 familiar and straightforward, particularly when using calculators or log tables designed for base 10 computations.This understanding is applied when solving our equation, explaining why we called upon the base frequently throughout the problem's solution.