Problem 56 - Intermediate Algebra

Question

Solve each equation. $$ \log _{2}(x-7)+\log _{2} x=3 $$

Step-by-Step Solution

Answer

The solution is \(x = 8\).

Step 1: Use Logarithmic Property

Recall that the property \(\log_a m + \log_a n = \log_a (mn)\) allows us to combine the two logarithmic expressions. Therefore, \(\log_{2}(x-7) + \log_{2}x = \log_{2}((x-7)x)\). This simplifies our equation to: \[ \log_{2}(x^2 - 7x) = 3 \]

Step 2: Use Definition of Logarithm

Using the definition of logarithm, \(\log_{b}(A)=C\) can be rewritten as \(b^C = A\). In this case, that means using \(\log_{2}(x^2 - 7x) = 3\), we convert this to: \[ 2^3 = x^2 - 7x \] \(2^3\) simplifies to 8, so we have: \[ x^2 - 7x = 8 \]

Step 3: Rearrange into Quadratic Equation

Rearrange the equation to set it equal to zero: \(x^2 - 7x - 8 = 0\). This is a standard quadratic equation.

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -7\), and \(c = -8\). Calculate the discriminant: \(b^2 - 4ac = (-7)^2 - 4\cdot1\cdot(-8) = 49 + 32 = 81\). Now solve for \(x\): \[ x = \frac{-(-7) \pm \sqrt{81}}{2 \cdot 1} = \frac{7 \pm 9}{2} \] Which gives us the solutions \(x = 8\) and \(x = -1\).

Step 5: Validate the Solutions

Check the solutions against the original equation. The expression \(x-7\) must be positive for the logarithm to be defined. Therefore, substitute:- If \(x = 8\), \log_{2}(8-7) + \log_{2}(8) = 3\; checks out as: \(\log_{2}(1) + \log_{2}(8) = 0 + 3 = 3\).- If \(x = -1\), \log_{2}(-1-7) + \log_{2}(-1) = 3\; here, both logarithms are undefined as they are calculated on negative numbers.Thus, only \(x = 8\) is valid.

Key Concepts

Logarithmic Properties

When working with logarithms, understanding the properties that govern them is crucial. For instance, one essential logarithmic property is that

• \( \log_a m + \log_a n = \log_a (mn) \)

This property is particularly helpful for simplifying expressions that involve the sum of two logarithms with the same base. It means you can combine the logs into one single log. In the initial problem, this property allowed for the expression \( \log_{2}(x-7) + \log_{2}x \) to be rewritten as \( \log_{2}((x-7)x) \), leading to the simpler equation \[ \log_{2}(x^2 - 7x) = 3 \]Another important logarithmic tool is transforming the log equation into an exponential form using the definition of a logarithm, which helps in converting logarithmic problems into algebraic ones.

• If \( \log_b(A) = C \), then it can be converted into \( b^C = A \).

In our example, this transformation helped convert the logarithmic form to an equation of the exponential form: \[ 2^3 = x^2 - 7x \]These properties allow you to manipulate and solve logarithmic expressions more effectively.

Quadratic Equations

Once the logarithmic portion of an equation is dealt with, algebraic manipulation often brings us to another type of equation known as a quadratic equation. The standard form of a quadratic equation is: \[ ax^2 + bx + c = 0 \]In our case, after simplifying, the equation \( x^2 - 7x = 8 \) was rearranged to the quadratic form: \[ x^2 - 7x - 8 = 0 \]Quadratic equations can have two solutions, which can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 1 \), \( b = -7 \), and \( c = -8 \). Inserting these values into the quadratic formula gives: \[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]This results in the solutions \( x = 8 \) and \( x = -1 \). Remember that quadratic equations, depending on their discriminant \( b^2 - 4ac \), can have two distinct real solutions, one real solution, or no real solutions. In this exercise, the discriminant \( 81 \) is positive, leading to two real and distinct solutions.

Validation of Solutions

After solving equations, one crucial final step is the validation of solutions to make sure they meet all conditions specified in the original problem. Special attention should be paid to any specific requirements of the mathematical expressions involved. In the original logarithmic equation, both log terms must have positive arguments.

• Logarithms are undefined for zero or negative numbers. Hence, any solution must ensure the argument of each logarithm remains positive.

For instance, substituting \( x = 8 \) back into the equation: \[ \log_{2}(8-7) + \log_{2}(8) = 3 \] results in \( \log_{2}(1) + \log_{2}(8) = 0 + 3 = 3 \). This substitution checks out, meaning \( x = 8 \) is valid. However, substituting \( x = -1 \) results in negative arguments for the logs, making these calculations undefined and, consequently, this solution invalid. Therefore, only \( x = 8 \) is acceptable in this context. Always verify that solutions comply with the initial conditions of the problem to ensure they are truly valid.