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Chapter 8: Problem 45

Solve each equation. $$ \log 5-\log 2 x=1 $$

Short Answer

Expert verified
The solution for the given logarithmic equation is \(x = \frac{1}{2}\).

Step by step solution

01

Use the properties of logarithms

The difference of two logarithms with the same base can be re-written as the logarithm of the quotient of the two numbers. Hence, we can rewrite the given equation \(\log 5 - \log 2x = 1\) as \(\log \frac{5}{2x} = 1\).
02

Convert the equation to exponential form

A logarithmic equation \( \log_b a = c\) can be rewritten as an exponential equation \(b^c = a\). So, \(\log \frac{5}{2x} = 1\) can be converted to the exponential form \(10^{1}=\frac{5}{2x}\).
03

Solve for x

First, express the left-hand side. \(10^{1} = 10\), so we get \(10 = \frac{5}{2x}\). Now, isolate \(x\) by multiplying both sides by \(2x\) and then dividing by 10: \(x=\frac{5}{10}\times2 = \frac{1}{2}\).

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Key Concepts

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

Properties of Logarithms
Logarithms have several properties that make them an essential tool in mathematics. One of these properties is the subtraction of logarithms. When you see an expression like \( \log a - \log b \), you can rewrite it as \( \log \frac{a}{b} \). This is known as the quotient rule.
  • This property allows you to simplify expressions and solve logarithmic equations more easily.
  • It's based on the idea that logarithms are exponents. Subtracting exponents (logs) is like dividing their corresponding numbers.
  • For instance, applying this to our original problem converts \( \log 5 - \log 2x \) into \( \log \frac{5}{2x} \).
Understanding how these properties work will allow you to unlock solutions to more complex problems.
Exponential Form
Switching between logarithmic and exponential forms is a handy trick in mathematics. When we have a logarithmic equation like \( \log_b a = c \), we can convert it into an exponential equation: \( b^c = a \).
  • This conversion is crucial because sometimes it's easier to solve an equation when it's in exponential form.
  • In our example, \( \log \frac{5}{2x} = 1 \) becomes \( 10^1 = \frac{5}{2x} \).
  • Using the base of common logarithms, which is 10, we see that this relationship simplifies calculations.
This transformation helps us find solutions more directly, especially when dealing with equations that involve unknowns.
Solving Equations
Solving equations requires finding the unknown value that satisfies the given condition. With our equation in exponential form, we can easily proceed to find the value of \( x \).
  • First, evaluate the left side: \( 10^1 = 10 \).
  • Set up the equation \( 10 = \frac{5}{2x} \) to isolate \( x \).
  • Multiply both sides by \( 2x \) to eliminate the fraction: \( 10 \times 2x = 5 \).
  • Divide by 10 to solve for \( x : x = \frac{5}{10} \times 2 \).
  • Finally, simplify to find \( x = \frac{1}{2} \).
By following these steps, we've determined the solution to the equation, using a mix of algebraic manipulation and understanding of exponential relationships.

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Most popular questions from this chapter

Write each equation in logarithmic form. \(5^{-3}=\frac{1}{125}\)

Use the Change of Base Formula to evaluate each expression. Then convert it to a logarithm in base \(8 .\) $$ \log _{2} 7 $$

Use the Change of Base Formula to evaluate each expression. Then convert it to a logarithm in base \(8 .\) $$ \log _{3} 54 $$

Use the Change of Base Formula to evaluate each expression. Then convert it to a logarithm in base \(8 .\) $$ \log _{3} 33 $$

Solve each equation. Check your answers. $$ 2 \log (x+1)=5 $$
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