Problem 87 - Precalculus

Question

Solve the equation: \(\log (x)+\log (x+3)=1\) for \(x\) Solution: Apply the product property (5). \(\quad \log \left(x^{2}+3 x\right)=1\) Exponentiate both sides (base 10 ). \(10^{\log \left(x^{2}+3 x\right)}=10^{1}\) Apply the property of inverses. \(x^{2}+3 x=10\) Factor. \(\quad(x+5)(x-2)=0\) Solve for \(x . \quad \quad x=-5\) and \(x=2\) This is incorrect. What mistake was made?

Step-by-Step Solution

Answer

The mistake was including \(x=-5\) as a solution; only \(x=2\) is valid.

Step 1: Apply the Product Property of Logarithms

We start by using the product property of logarithms, which states that \(\log(a) + \log(b) = \log(a \times b)\). In the given equation, this lets us rewrite \(\log(x) + \log(x+3) = \log(x(x+3)) = 1\).

Step 2: Exponentiate to Eliminate the Logarithm

To remove the logarithm, exponentiate both sides of the equation with base 10. This results in \(10^{\log(x(x+3))} = 10^1\). The logarithmic function and exponential function are inverses, so \(x(x+3) = 10\).

Step 3: Rearrange into a Standard Quadratic Equation

Expand the left side to get \(x^2 + 3x = 10\). Rearrange this into a standard quadratic form: \(x^2 + 3x - 10 = 0\).

Step 4: Factor the Quadratic Equation

The quadratic equation can be factored into \((x+5)(x-2) = 0\).

Step 5: Solve for x

Set each factor in the factored equation equal to zero: \(x+5=0\) and \(x-2=0\). Solving these gives \(x=-5\) and \(x=2\).

Step 6: Check the Solutions

Substitute \(x=-5\) and \(x=2\) back into the original equation \(\log(x) + \log(x+3) = 1\). \(x=-5\) results in negative values inside the logarithms, which are undefined. Thus, \(x=-5\) is not a valid solution. \(x=2\) yields valid logarithmic expressions, so it is the only valid solution.

Key Concepts

Product Property of Logarithms

The product property of logarithms is a key tool in simplifying expressions. This property states that

• \(\log(a) + \log(b) = \log(a \times b)\)

This means if you have the sum of two logs with the same base, you can combine them into a single log by multiplying their arguments.
In our original exercise, we applied this property to \(\log(x) + \log(x+3)\). By the product property, this becomes \(\log(x \times (x + 3))\), which simplifies to \(\log(x^2 + 3x)\).
Using properties like these can streamline equations, providing a handy shortcut when dealing with logarithmic expressions.

Quadratic Equation

Quadratic equations are expressions that take the form of \(ax^2 + bx + c = 0\). Here,

• \(a\), \(b\), and \(c\) are constants,
• \(x^2\) is the variable raised to the second power.

In our exercise, after applying the product property and exponentiating, we found ourselves with the equation \(x^2 + 3x = 10\). To transform this into the standard quadratic form, we rearranged it as \(x^2 + 3x - 10 = 0\).
Quadratic equations are a fundamental algebraic structure. Solving them is a valuable skill, often requiring further methods like factoring or using the quadratic formula.

Factoring

Factoring is a technique used to break down an expression into simpler terms, or factors, that multiply together to give the original expression. It is especially useful for solving quadratic equations. In our example,

• We started with \(x^2 + 3x - 10 = 0\).
• Through factoring, this quadratic splits into \((x + 5)(x - 2) = 0\).

Each factor corresponds to a potential solution when set equal to zero. For our quadratic, solving \(x + 5 = 0\) gives \(x = -5\), and \(x - 2 = 0\) gives \(x = 2\).
However, always verify these solutions in the context of the original problem, as only \(x = 2\) is valid in this case, since logarithms of negative numbers are undefined.