Problem 48 - Algebra and Trigonometry

Question

Solve the logarithmic equation for \(x\) $$ \log x+\log (x-3)=1 $$

Step-by-Step Solution

Answer

The solution is \(x = 5\).

Step 1: Apply Logarithmic Identity

We start by applying the logarithmic identity \( ext{log} (a) + ext{log} (b) = ext{log} (ab)\). Using this identity, we can combine the two logarithms on the left side of the equation as follows:\[\log (x) + \log (x-3) = \log (x(x-3))\] This simplifies our equation to:\[\log(x(x-3)) = 1\]

Step 2: Convert to Exponential Form

To remove the logarithm, we convert the equation from logarithmic to exponential form. If \(\log (y) = a\), then \(y = 10^a\). Applying this to our equation, we get:\[x(x-3) = 10^1\]Which simplifies to:\[x(x-3) = 10\]

Step 3: Expand and Rearrange the Equation

Expand the equation \(x(x-3) = 10\) to get:\[x^2 - 3x = 10\]Rearrange it to form a standard quadratic equation:\[x^2 - 3x - 10 = 0\]

Step 4: Solve the Quadratic Equation

We will use the quadratic formula to solve for \(x\), where the quadratic equation is of the form \(ax^2 + bx + c = 0\). Here, \(a = 1\), \(b = -3\), and \(c = -10\). The quadratic formula is:\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]Substitute the values into the formula:\[x = \frac{{3 \pm \sqrt{{(-3)^2 - 4 \cdot 1 \cdot (-10)}}}}{2 \cdot 1}\]\[x = \frac{{3 \pm \sqrt{{9 + 40}}}}{2}\]\[x = \frac{{3 \pm \sqrt{49}}}{2}\]\[x = \frac{{3 \pm 7}}{2}\]This gives two potential solutions:1. \(x = \frac{{3 + 7}}{2} = 5\)2. \(x = \frac{{3 - 7}}{2} = -2\)

Step 5: Verify the Solutions and Exclude Non-Domain Values

The solution \(x = -2\) is not valid since we cannot take the logarithm of a negative number or zero. Thus, only \(x = 5\) is in the domain of the original equation. We verify this by substitution:\[\log 5 + \log (5-3) = \log 5 + \log 2 = \log (5 \times 2) = \log 10 = 1\]Therefore, \(x = 5\) satisfies the original equation.

Key Concepts

Logarithmic Identity

When working with logarithmic equations, a common tool you'll use is the logarithmic identity. This identity states that the sum of two logarithms can be expressed as a single logarithm by multiplying their arguments. In mathematical terms, if you have

• \( \log(a) + \log(b) \)

then it becomes

• \( \log(ab) \).

This is particularly useful in simplifying equations by reducing the number of logarithmic terms. In the problem given, you applied this identity to convert

• \( \log(x) + \log(x-3) \)
• into \( \log(x(x-3)) \),

thus simplifying your equation significantly.
It is essential to ensure the domain of the original expressions are considered valid throughout the calculations. With logarithmic expressions, avoid taking the logarithm of a negative number or zero, as they are not within the domain of the logarithmic function.

Quadratic Formula

The quadratic formula is a standard method used to find the solutions of quadratic equations, which are polynomials of the form

• \( ax^2 + bx + c = 0 \).

The formula is expressed as:

• \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]

You use this formula when the quadratic equation cannot be easily factored. Each part of the equation contributes to solving for the values of \( x \). Here, \( a \) is the coefficient of \( x^2 \), \( b \) is the coefficient of \( x \), and \( c \) is the constant term. In our exercise, the quadratic equation derived from converting the logarithmic equation was

• \( x^2 - 3x - 10 = 0 \),

with coefficients \( a = 1 \), \( b = -3 \), and \( c = -10 \). By substituting these into the quadratic formula, we determined the potential solutions for \( x \). Remember, once you find these solutions, check them against the original equation to ensure they are valid considering the problem's domain restrictions.

Exponential Form Conversion

Converting logarithmic equations to exponential form is a key step in many problems. This conversion uses the idea that if you have

• \( \log_b(y) = a \),

then it can be expressed in exponential form as

• \( y = b^a \).

The exponential form translates the logarithmic statement into a multiplication form, which often makes it easier to solve for unknowns. In this exercise, after applying the logarithmic identity, the equation was simplified to

• \( \log(x(x-3)) = 1 \).

To express it in exponential form, we treat the basis of the logarithm as 10 (commonly used for common logarithms) leading to

• \( x(x-3) = 10^1 \) or \( x(x-3) = 10 \).

By doing this, the equation becomes a quadratic equation, allowing for further solution using the quadratic formula. Transitioning between these forms is instrumental in efficiently marching towards the problem's solution.