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Chapter 4: Problem 26

Each of these equations involves more than one logarithm. Solve each equation. Give exact solutions. $$\log _{3}(x)=\log _{3}(2)-\log _{3}(x-2)$$

Short Answer

Expert verified
The exact solution is \( x = 1 + \sqrt{3} \).

Step by step solution

01

Simplify the Logarithmic Equation

Use the properties of logarithms to combine the right side of the equation. Recall that \ \ \(\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)\) \ \ The equation is: \ \ \( \log_{3}(x) = \log_{3}(2) - \log_{3}(x-2) \) \ \ Combine the right side: \ \ \( \log_{3}(x) = \log_{3} \left( \frac{2}{x-2} \right) \)
02

Set the Arguments Equal

Since the logarithms are equal, their arguments must also be equal. \ \ \(x = \frac{2}{x-2} \)
03

Solve the Equation

Solve the equation by isolating \(x\): \ \ \( x (x - 2) = 2 \) \ \ Expand and simplify: \ \ \( x^2 - 2x = 2 \) \ \ \( x^2 - 2x - 2 = 0 \)
04

Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) : \ \ Here, \(a = 1\), \(b = -2\), and \(c = -2\). \ \ Calculate the discriminant: \ \ \( \Delta = (-2)^2 - 4(1)(-2) = 4 + 8 = 12 \) \ \ So, \ \ \( x = \frac{2 \pm \sqrt{12}}{2} \) \ \ Simplify: \ \ \( x = \frac{2 \pm 2\sqrt{3}}{2} \) \ \ \( x = 1 \pm \sqrt{3} \)
05

Verify the Solutions

Verify if the solutions fit the original equation. \ \ For \( x = 1 + \sqrt{3} \): \ \ \(x > 2 \), so this solution is valid. \ \ For \( x = 1 - \sqrt{3} \): \ \ \( x < 2 \), so this solution is not valid as we cannot take the logarithm of a negative number. \ \ The valid solution is: \( x = 1 + \sqrt{3} \)

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Key Concepts

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

Properties of Logarithms
To solve logarithmic equations, understanding the properties of logarithms is essential. These properties simplify complex logarithmic expressions and help make equations more manageable.
  • Product Property: \(\text{log}_b(MN) = \text{log}_b(M) + \text{log}_b(N)\)
  • Quotient Property: \(\text{log}_b\bigg(\frac{M}{N}\bigg) = \text{log}_b(M) - \text{log}_b(N)\)
  • Power Property: \(\text{log}_b(M^p) = p \text{log}_b(M)\)

In our exercise, we used the quotient property to combine the right-side logarithmic terms:
\(\text{log}_3(2) - \text{log}_3(x - 2) = \text{log}_3\bigg(\frac{2}{x-2}\bigg)\). This step drastically reduces the complexity of our equation.
Quadratic Equations
Once we simplified the logarithmic equation, it transformed it into a quadratic form: \(\text{log}_3(x) = \text{log}_3\bigg(\frac{2}{x-2}\bigg)\) became \(\frac{2}{x-2} = x \). Solving it gives us
  • \( x(x-2) = 2 \)
  • Expanding:
    \(x^2 - 2x = 2\)
  • Rearranging into standard form:
    \(x^2 - 2x - 2 = 0\)

Recognize this as a standard quadratic equation: ax^2 + bx + c = 0, with a=1, b=-2, c=-2. To solve this, we apply the quadratic formula:
\(\text{quadratic\text{ formula:} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Solving for Variables
Solving for variables involves isolating the unknown variable on one side of the equation. After forming the quadratic equation \ (x^2 - 2x - 2 = 0) \, use the quadratic formula:
  • Here: a=1, b=-2, c=-2
  • Calculate discriminant: \(\triangle = b^2 - 4ac = 4 + 8 = 12\)
  • So, \ (x = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3})\
Now, we have potential solutions: \ (x = 1 + \sqrt{3})\ and \ (x = 1 - \sqrt{3}).\
However, verify both solutions by substituting them back into the original equation.
If a solution does not make logical sense (like a negative logarithmic argument), disregard it. In this case, \ (x = 1 \pm \sqrt{3})\ provides two nominal solutions, but only \ (1 + \sqrt{3})\ is valid because it's greater than 2.
So, the valid solution to the equation is \ (1 + \sqrt{3}).\

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Most popular questions from this chapter

To evaluate an exponential or logarithmic function we simply press a button on a calculator. But what does the calculator do to find the answer? The next exercises show formulas from calculus that are used to evaluate \(e^{x}\) and \(\ln (1+x)\). Infinite Series for \(e^{x}\) The following formula from calculus is used to compute values of \(e^{x}\) : $$e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\cdots+\frac{x^{n}}{n !}+\cdots$$ where \(n !=1 \cdot 2 \cdot 3 \cdot \cdots \cdot n\) for any positive integer \(n .\) The notation \(n !\) is read " \(n\) factorial." For example, \(3 !=1 \cdot 2 \cdot 3=6\) In calculating \(e^{x},\) the more terms that we use from the formula, the closer we get to the true value of \(e^{x}\). Use the first five terms of the formula to estimate the value of \(e^{0.1}\) and compare your result to the value of \(e^{0.1}\) obtained using the \(e^{x}-\) key on your calculator.

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