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Chapter 4: Problem 21

Solve the equation. $$\log x-\log (x+1)=3 \log 4$$

Short Answer

Expert verified
No solution in the real number domain.

Step by step solution

01

Use Logarithm Properties

Recall that the difference of two logarithms, \(\log a - \log b\), can be written as \(\log \left(\frac{a}{b}\right)\). Apply this property to transform the given equation:\[\log x - \log (x+1) = \log \left(\frac{x}{x+1}\right)\] Thus, the equation becomes:\[\log \left(\frac{x}{x+1}\right) = 3 \log 4\]
02

Apply Another Logarithm Property

Recall that \(n \log a = \log (a^n)\). Use this property for \(3 \log 4\):\[3 \log 4 = \log (4^3) = \log 64\] So, the equation transforms to:\[\log \left(\frac{x}{x+1}\right) = \log 64\]
03

Equate the Arguments

Since the logarithms are equal on both sides of the equation, their arguments must also be equal. Set the arguments equal to each other:\[\frac{x}{x+1} = 64\]
04

Solve the Equation

Solve the equation \(\frac{x}{x+1} = 64\) by cross-multiplying:\[x = 64(x + 1)\]Expand and simplify the equation:\[x = 64x + 64\]Rearrange the terms to isolate \(x\) on one side:\[0 = 63x + 64\]\[63x = -64\]Divide both sides by 63 to solve for \(x\):\[x = -\frac{64}{63}\]
05

Check the Solution for Validity

Substitute \(x = -\frac{64}{63}\) into the original logarithmic equation to ensure it satisfies the conditions. Since logarithms of negative numbers are undefined, this solution \(x = -\frac{64}{63}\) does not satisfy the equation. Hence, there is no valid solution for this equation in the real number domain.

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Key Concepts

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

Logarithm Properties
When tackling logarithmic equations, understanding the fundamental properties of logarithms is crucial. One important property is the difference of two logs, expressed as \(\log a - \log b = \log \left(\frac{a}{b}\right)\). This allows you to condense expressions and simplify equations.
  • The property is helpful for turning subtractions into divisions within a logarithm.
  • It makes solving equations straightforward by minimizing the terms you are dealing with.
Another key property deals with "power" multiplication, represented by \(n \log a = \log (a^n)\). This lets you simplify expressions involving coefficients multiplying the log.
  • This property is useful when handling equations with coefficients in front of log terms.
  • By transforming such terms, you can equate the arguments of the logs.
Understanding and applying these properties enables you to manage logarithmic equations with confidence.
Exponential Functions
Exponential functions often come into play when you convert logarithmic equations into an exponential form. Consider the equation \(\log \left(\frac{x}{x+1}\right) = \log 64\). Here, both the left and right side are logarithms.
  • This signals that the arguments, \(\frac{x}{x+1}\) and 64, must be equal because the logarithms of equal quantities are themselves equal.
  • The mismatch of arguments is resolved by equating \(\frac{x}{x+1} = 64\).
Exponential functions can also arise when transforming a logarithmic form into an exponential one. This can sometimes simplify solving the equation, giving a clearer path to the solution.
Feel free to lean on exponential functions when working with logarithms, as they often go hand in hand, turning complex logarithmic forms into more tangible exponential ones.
Equation Solving
The art of solving equations involves transforming and simplifying them step by step. When given the equation \(\frac{x}{x+1} = 64\), this needs direct handling:
  • First, use cross-multiplication to eliminate the fraction.
  • This gives \(x = 64(x + 1)\).
  • Expand and simplify by distributing \(64\) into \(x+1\).
By rearranging terms, you isolate the variable. This means moving all terms with \(x\) on one side and constants on the other, leading to:
  • \(x = 64x + 64\)
  • Simplify further, \(0 = 63x + 64\).
  • Get \(63x = -64\) and divide by \(63\) to find \(x\).
Solid understanding of basic algebraic manipulation is integral to solving logarithmic and nonlinear equations effectively.
Real Number Domain
The real number domain refers to all possible values that a variable can take on a continuous number line, excluding complex or imaginary numbers.
  • When working with logarithms, it's essential to remember that logarithm inputs must always be positive real numbers.
  • Any solution making the argument of a logarithm negative or zero is inadmissible.
In the equation \(\log x - \log (x+1)=3 \log 4\), the candidate solution was \(x = -\frac{64}{63}\). However, substituting this back reveals the input value is negative.
  • This means the solution falls outside the allowable real number domain for logarithms.
Thus, there are specific considerations, such as ensuring positive logarithm arguments, you must keep in mind when evaluating solutions within the real number domain for equations like these.

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Most popular questions from this chapter

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