Problem 533 - Precalculus

Question

Use the one-to-one property of logaritims to find an exact solution for \(\log _{8}(7)+\log _{8}(-4 x)=\log _{8}(5) .\) If there is no solution, write no solution.

Step-by-Step Solution

Answer

The exact solution for the equation is \( x = -\frac{5}{28} \).

Step 1: Apply the Product Rule of Logarithms

The task is to combine the logs on the left side of the equation by using the product rule of logarithms, which states \( \log_b(a) + \log_b(c) = \log_b(ac) \). Therefore, \( \log_{8}(7) + \log_{8}(-4x) = \log_{8}(7 imes (-4x)) \), simplifying to \( \log_{8}(-28x) \). The equation now reads \( \log_{8}(-28x) = \log_{8}(5) \).

Step 2: Use the One-to-One Property

The one-to-one property of logarithms states that if \( \log_b(a) = \log_b(c) \), then \( a = c \). Applying this property allows us to set the arguments equal: \( -28x = 5 \).

Step 3: Solve for x

To solve \( -28x = 5 \), divide both sides of the equation by \(-28\) to isolate \( x \). This results in \( x = -\frac{5}{28} \).

Key Concepts

Product Rule of Logarithms

The product rule of logarithms is a key part of simplifying expressions involving logarithms. It is a rule that partly makes logarithms powerful tools when dealing with multiplication inside log expressions. The product rule states that the logarithm of a product is the same as the sum of the logarithms of the factors. In mathematical terms, this is expressed as: \ \[ \log_b(a) + \log_b(c) = \log_b(ac) \]\ Applying this rule allows us to turn expressions of logs added together into a single logarithm containing a multiplied argument. For example, in the problem \( \log_{8}(7) + \log_{8}(-4x) \), using the product rule simplifies it to \( \log_{8}(7 \times (-4x)) \), which is \( \log_{8}(-28x) \). This simplification can greatly help in solving logarithmic equations since it reduces the number of log terms and facilitates easier comparison and operations.

Solving Logarithmic Equations

Solving logarithmic equations often involves applying various logarithm properties and algebraic techniques. Once you express logarithms in a simpler form, such as employing the product rule, you can use other methods like the one-to-one property. This property is essential in solving equations like \( \log_b(a) = \log_b(c) \). According to the one-to-one property, if both sides of the equation are logs with the same base, then their arguments must be equal. This allows you to equate and solve the arguments directly: \ \[ a = c \]\ For example, with the equation \( \log_{8}(-28x) = \log_{8}(5) \), applying the one-to-one property gives us \( -28x = 5 \). This new equation involves no logarithms and can be solved using standard algebraic manipulation. Solving for \( x \) involves isolating \( x \) by division, yielding \( x = -\frac{5}{28} \). Such techniques are invaluable tools for transforming complex logarithmic equations into more solvable forms.

No Solution in Logarithms

In certain logarithmic equations, it's possible to reach a point where no real solution exists. This scenario typically arises when the arguments of the logarithms result in a mathematical impossibility. For the logarithm to be valid, the argument must be positive because the logarithm of any non-positive number is undefined within the realm of real numbers. Consider an equation that solves into something like \( \log_{b}(c) = \log_{b}(-a) \). Here, if \(-a\) implies a negative argument, such a logarithm can’t exist for any real number base \(b\). In the original exercise, although the variable \( x \) turned negative, the argument \(-4x\) still resulted in a positive number when multiplied by seven, thus maintaining a valid input for the logarithm. However, if solving leads to a negative argument, we acknowledge with "no solution," because the log of a negative argument doesn't exist.