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Chapter 4: Problem 75

Solve each equation and check your answers. $$\log (2 x-5)-\log 78=-1$$

Short Answer

Expert verified
The solution is \( x = 6.4 \).

Step by step solution

01

Use Logarithm Property

Identify and apply the logarithm property: \( \log a - \log b = \log \frac{a}{b} \). Therefore, combine the logarithms into one: \( \log \frac{2x-5}{78} = -1 \).
02

Eliminate the Logarithm

Recognize that solving the equation \( \log b = -1 \) implies that \( b = 10^{-1} \). Apply this to the equation: \( \frac{2x-5}{78} = 0.1 \).
03

Solve for \( x \)

Multiply both sides by 78 to eliminate the fraction: \( 2x - 5 = 7.8 \). Then add 5 to both sides to isolate the term with \( x \): \( 2x = 12.8 \). Finally, divide by 2 to solve for \( x \): \( x = 6.4 \).
04

Check Your Solution

Substitute \( x = 6.4 \) back into the original equation: \( \log(2(6.4) - 5) - \log 78 = -1 \). Simplify inside the log: \( \log(12.8 - 5) - \log 78 = -1 \), which results in \( \log(7.8) - \log 78 = -1 \). The final calculation \( \log \frac{7.8}{78} \) does indeed equal \( -1 \), confirming that \( x = 6.4 \) is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Properties
When dealing with logarithmic equations, understanding the properties of logarithms is essential. One of the most useful properties is the Difference of Logs, which states:
  • \( \log a - \log b = \log \frac{a}{b} \)
This property allows us to combine two separate logarithmic terms into a single logarithmic expression, making the equation easier to solve. In the given problem, the use of this property transforms \( \log (2x-5) - \log 78 = -1 \) into \( \log \frac{2x-5}{78} = -1 \). Understanding and applying these properties can immensely simplify solving logarithmic equations.
Solving Equations
Once the logarithmic expression is simplified using properties, the next step is to solve the equation. The goal is to eliminate the logarithm to find the value of \( x \). In the exercise, the equation becomes \( \log \frac{2x-5}{78} = -1 \).To remove the log, utilize the characteristic of logarithms that if \( \log b = n \), then \( b = 10^n \). Hence, for our equation, \( \frac{2x-5}{78} = 10^{-1} = 0.1 \). Solving this linear equation involves basic algebra:
  • First, multiply both sides by 78 to eliminate the fraction: \( 2x-5 = 7.8 \).
  • Next, add 5 to both sides: \( 2x = 12.8 \).
  • Finally, divide by 2: \( x = 6.4 \).
By following these steps, the solution for \( x \) is achieved efficiently.
Check Solution
Verifying your solution is a crucial step in solving equations. It ensures that the solution satisfies the original equation. In this context, check whether \( x = 6.4 \) works in the original logarithmic equation.Substitute \( x = 6.4 \) back into the equation:
  • Calculate \( 2(6.4)-5 = 7.8 \).
  • The equation becomes \( \log(7.8) - \log(78) = -1 \).
  • This simplifies to \( \log \frac{7.8}{78} = \log(0.1) = -1 \).
As the logarithmic expressions match, the solution is confirmed as correct. Checking your answers, especially in precalculus, reinforces the concepts and guarantees accuracy.
Precalculus Practice
Logarithmic equations form an essential part of precalculus, as they lay the groundwork for more advanced mathematical topics in calculus and beyond. Mastering these equations involves not only applying algebraic skills but also a sound understanding of logarithmic properties and logical reasoning. In precalculus practice, frequent exercises with problems like the one provided help to solidify these foundational skills. Regular practice allows you to:
  • Become comfortable with converting and simplifying logarithmic expressions.
  • Gain expertise in solving equations through systematic approaches.
  • Develop a habitual checking process to avoid mistakes and confirm your solutions.
Engaging with these concepts repeatedly ensures you're prepared for the challenges of calculus and helps nurture mathematical intuition.

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