Problem 71 - Algebra and Trigonometry

Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. $$ \text { Solve: } \log (x+1)+\log (x-2)=1 $$

Step-by-Step Solution

Answer

The solution is x = 4.

Step 1: Apply Logarithm Properties

Use the properties of logarithms to combine the logarithmic expressions. Recall that \(\log_a b + \log_a c = \log_a (bc)\). Applying this property gives: \[ \log((x+1)(x-2)) = 1\]

Step 2: Convert Logarithmic to Exponential Form

Rewrite the equation in exponential form. Recall that \(\log_b a = c \) is equivalent to \(a = b^c\). Thus, \[ (x+1)(x-2) = 10^1 \] which simplifies to: \[ (x+1)(x-2) = 10 \]

Step 3: Expand the Equation

Expand the product on the left-hand side: \[ x^2 - 2x + x - 2 = 10 \] Simplify this to get a standard quadratic equation: \[ x^2 - x - 2 = 10 \]

Step 4: Arrange into Standard Quadratic Form

Rearrange the equation into standard quadratic form: \[ x^2 - x - 12 = 0 \]

Step 5: Solve the Quadratic Equation

Factor the quadratic equation. Find two numbers that multiply to -12 and add to -1. These numbers are -4 and 3: \[ (x-4)(x+3) = 0 \] Set each factor to zero and solve for \(x\): \[ x-4 = 0 \] \[ x = 4 \] and \[ x+3 = 0 \] \[ x = -3 \]

Step 6: Verify Solutions

Verify that the solutions satisfy the original logarithmic equation. Substitute \(x = 4\) into the original equation: \[ \log(4+1) + \log(4-2) = \log 5 + \log 2 = \log(5 \cdot 2) = \log 10 = 1 \] Thus, \(x = 4\) is a valid solution. However, substitute \(x = -3\) into the original equation: \[ \log(-3+1) + \log(-3-2) \] The terms \(\log(-2)\) and \(\log(-5)\) are undefined since the log of a negative number is not defined in the real numbers. Therefore, \(x = -3\) is not a valid solution.

Key Concepts

Logarithm Properties

Logarithms can be tricky, but they have some useful properties that make solving equations easier. One important property is the product rule: \[\text{log}_a(b) + \text{log}_a(c) = \text{log}_a(bc)\]. This means you can combine two logarithms with the same base into one by multiplying their arguments. In our exercise, we used this rule to combine \[\text{log}(x+1) + \text{log}(x-2)\] into \[\text{log}((x+1)(x-2))\]. Understanding and applying these rules can simplify complex logarithmic expressions, making them easier to solve. Remember, logarithms also have other important properties such as the quotient and power rules, which can help in different scenarios.

Exponential Form

Sometimes, solving logarithmic equations involves converting them to exponential form. The idea is to remember that \[\text{log}_b(a) = c\] is the same as \[\text{a} = b^c\]. In our exercise, we used this property to change \[\text{log}((x+1)(x-2)) = 1\] into \[(x+1)(x-2) = 10^1\]. Converting to exponential form often simplifies the equation because exponents can be easier to manipulate and solve directly. This conversion step is crucial for transforming a logarithmic equation into an algebraic equation that is more familiar and easier to handle.

Quadratic Equation

After converting the logarithmic equation to exponential form, we ended up with a quadratic equation. A standard quadratic equation looks like \[{ax}^2 + bx + c = 0\]. In our example, expanding and simplifying led to \[{x}^2 - x - 12 = 0\]. Solving quadratic equations often involves factoring, but you can also use the quadratic formula: \[-b \pm \sqrt{b^2 - 4ac} / 2a\]. In this case, we factored it to \[(x-4)(x+3) = 0\], giving solutions x=4 and x=-3. However, always verify solutions back in the original equation. Some roots, like \[-3\] here, can make the original logarithm undefined, so those are discarded. This verification step ensures that only valid solutions are considered.