Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{3} x+\log _{3}(x-8)=2\)

Short answer

The solution to the equation \( \log_3 x + \log_3 (x - 8) = 2 \) is \( x = 9 \).

Step-by-step solution

Use the Product Rule

The first step is to combine the two logarithms into one using the product property of logarithms, which states that \( \log_b (MN) = \log_b M + \log_b N \). Therefore, we write \( \log_3 x + \log_3 (x - 8)= 2 \) as \( \log_3 [x(x - 8)] = 2 \).

Convert the Logarithm to an Exponential

Next, we use the logarithmic property \( \log_b a = c \) can be rewritten as \( b^c = a \) to convert our logarithm to an exponential. Thus, \( \log_3 [x(x - 8)] = 2 \) can be rewritten as \( 3^2 = x(x - 8) \).

Simplify the Equation

We now simplify \( 3^2 = x(x - 8) \) to \( 9 = x^2 - 8x \)

Rearrange to Quadratic Form

To solve for x, we first rearrange the equation into standard quadratic form. Therefore, \( 9 = x^2 - 8x \) rearranges to \( x^2 - 8x - 9 = 0 \).

Solve the Quadratic Equation

From here, we solve for x using the quadratic formula \( x = [-(-8) ± \sqrt{(-8)^2 - 4*1*(-9)}]/ (2*1) \), the solution for \( x^2 - 8x - 9 = 0 \) gives two possible values for x: \( x = 9 \) and \( x = -1 \).

Verify the Solutions

In a logarithmic equation, any value that makes the argument of the logarithm negative or zero is extraneous and must be discarded, since the logarithm of a negative number or zero is undefined. Therefore, the only solution to the original equation is \( x = 9 \), since \( x = -1 \) would make the argument of the second logarithm negative.

Key concepts

Product Rule of Logarithms

The Product Rule of Logarithms is a fundamental property that enables you to simplify expressions involving the sum of two logarithms with the same base. It states that the logarithm of a product is equal to the sum of the logarithms of its factors:

• \( \log_b (MN) = \log_b M + \log_b N \)

So, when you have an expression like \( \log_3 x + \log_3 (x - 8) \), it can be combined using the product rule to become \( \log_3 [x(x - 8)] \).
The ability to combine terms in this way is very useful for simplifying equations and making them easier to solve. It allows you to reduce multiple logarithmic terms into a single term, which can then be manipulated more easily through further algebraic steps. Understanding this rule is crucial when dealing with more complex logarithmic equations.

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) represents an unknown variable. Solving quadratic equations often involves finding the values of \( x \) that make the equation true. One way to solve quadratic equations is by using the quadratic formula:

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

In the given problem, after applying the product rule and converting to exponential form, you end up with a quadratic equation like \( x^2 - 8x - 9 = 0 \). This is a standard quadratic form, and to find the solution, you can apply the quadratic formula.
The calculation gives two potential solutions: \( x = 9 \) and \( x = -1 \). However, in the context of logarithmic equations, not all solutions will be valid. Any solution that makes the logarithm's argument negative or zero must be discarded, since logarithms are only defined for positive values.

Exponential Form of Logarithms

The exponential form of logarithms is a technique used to solve logarithmic equations by rewriting them as exponential equations. To convert a logarithmic equation \( \log_b a = c \) to its exponential form, you use:

• \( b^c = a \)

This step is crucial in solving the logarithmic equation: once you've combined the logarithms using the product rule, you transform \( \log_3 [x(x-8)] = 2 \) into its exponential form \( 3^2 = x(x-8) \).
This conversion simplifies the problem by removing the logarithm and producing a polynomial or quadratic equation that can be solved using standard algebraic techniques. Understanding this translation between logarithmic and exponential forms is key to efficiently tackling and solving logarithmic equations. It offers a clearer path to finding solutions by working within a numerical framework rather than the sometimes abstract logarithmic one.