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Chapter 4: Problem 47

Solve the logarithmic equation for \(x\). $$\log _{5}(x+1)-\log _{5}(x-1)=2$$

Short Answer

Expert verified
The solution is \(x = \frac{13}{12}\).

Step by step solution

01

Apply Logarithmic Identity

To start, use the logarithmic identity \( \log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right) \). This simplifies our equation to: \[ \log_5 \left( \frac{x+1}{x-1} \right) = 2 \]
02

Convert Logarithmic Equation to Exponential Form

Next, rewrite the equation in its exponential form. Remember the form \( \log_b(A) = C \) is equivalent to \( A = b^C \). So: \[ \frac{x+1}{x-1} = 5^2 \] This simplifies to: \[ \frac{x+1}{x-1} = 25 \]
03

Solve the Resulting Equation

Now solve the equation \( \frac{x+1}{x-1} = 25 \). Cross-multiply to get: \[ x+1 = 25(x-1) \] Then expand and simplify: \[ x+1 = 25x - 25 \] Bring all terms involving \(x\) to one side and constants to the other: \[ 1 + 25 = 25x - x \] \[ 26 = 24x \]
04

Solve for x

Divide both sides of the equation by 24 to solve for \(x\): \[ x = \frac{26}{24} \] Simplify the fraction: \[ x = \frac{13}{12} \]
05

Verify the Solution

Lastly, substitute \( x = \frac{13}{12} \) back into the original logarithmic equation to verify: \[ \log_5\left(\frac{13}{12} + 1\right) - \log_5\left(\frac{13}{12} - 1\right) = 2 \] Simplify inside the logarithms: \[ \log_5\left(\frac{25}{12}\right) - \log_5\left(\frac{1}{12}\right) = \log_5\left(25\right) \] \[ \log_5(25) = 2 \] The solution is verified.

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Key Concepts

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

Logarithmic Identities
Logarithmic identities are powerful tools in solving logarithmic equations. They allow us to simplify expressions into more manageable forms. In this problem, we have to work with the identity \( \log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right) \). This identity helps combine two separate logs, making the equation less complex. In our exercise, by applying this identity, we reduce \( \log_5(x+1) - \log_5(x-1) \) to \( \log_5\left(\frac{x+1}{x-1}\right) \). This crucial step transforms the problem into a simpler logarithmic equation that can be tackled with other techniques.
Exponential Form
When dealing with logarithms, converting an equation to exponential form often reveals the path to the solution. The log equation \( \log_b(A) = C \) can be rewritten as \( A = b^C \). This conversion is essential as it swaps the logarithm for an exponent, providing a new perspective in solving the equation. In this task, \( \log_5\left(\frac{x+1}{x-1}\right) = 2 \) translates to \( \frac{x+1}{x-1} = 5^2 \) or \( \frac{x+1}{x-1} = 25 \). This form is straightforward and sets the stage for algebraic manipulation by providing clear numbers to work with.
Cross-Multiplication
Cross-multiplication is a straightforward algebraic technique used in solving equations involving fractions. Once the logarithmic equation is in exponential form, like \( \frac{x+1}{x-1} = 25 \), you can cross-multiply to remove the fraction, simplifying the solve process. This involves multiplying both sides of the equation by the denominator of the fraction on the left side. This yields a linear equation: \( x+1 = 25(x-1) \). Expanding and simplifying, you get \( x+1 = 25x - 25 \), which then further simplifies to \( 26 = 24x \). Now, you have a simple equation that can be easily solved for \( x \).
Verify Solutions
After obtaining a solution, it is essential to verify it by plugging it back into the original equation. This checks if the solution satisfies the initial conditions given by the problem. In this case, substituting \( x = \frac{13}{12} \) into the original logarithmic equation ensures it holds true. This substitution leads to \( \log_5\left(\frac{25}{12}\right) - \log_5\left(\frac{1}{12}\right) = 2 \), verifying that the solution works since \( \log_5(25) \) also equals 2. This final step affirms that the value of \( x \) found is correct and satisfies the logarithmic equation.

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