Solving Logarithmic Equations Log Base 5 Of (10x-1) = Log Base 5 Of (9x+7)
In this article, we will delve into the process of solving a logarithmic equation. Logarithmic equations are equations that involve logarithms of expressions containing variables. These types of equations appear frequently in mathematics, physics, engineering, and various other fields. Understanding how to solve them is a fundamental skill for anyone working with quantitative problems. Here, we aim to provide a comprehensive explanation of how to solve the specific logarithmic equation: $\log _5(10 x-1)=\log _5(9 x+7)$. We will break down each step, providing clear reasoning and mathematical justifications along the way. By the end of this article, you will not only understand the solution to this particular problem but also gain a broader understanding of the methods used to solve logarithmic equations in general.
The key to solving logarithmic equations lies in understanding the properties of logarithms and how they interact with exponential functions. Logarithms are, in essence, the inverse operations of exponentiation. The equation $\log_b a = c$ is equivalent to $b^c = a$. This fundamental relationship allows us to convert logarithmic equations into more manageable algebraic forms. Another critical property is that if $\log_b x = \log_b y$, then $x = y$, provided that $x$ and $y$ are both positive and $b > 0$, $b \ne 1$. This property is crucial for simplifying equations where logarithms with the same base are set equal to each other. Furthermore, we must always check our solutions to ensure that they do not result in taking the logarithm of a negative number or zero, as logarithms are only defined for positive arguments. This step is vital in preventing extraneous solutions, which are values that satisfy the transformed equation but not the original logarithmic equation.
Understanding Logarithmic Equations
Before diving into the solution, let's clarify what logarithmic equations are and why they require a specific approach. A logarithmic equation is an equation that contains logarithms with variables. These equations arise in various contexts, such as calculating the intensity of earthquakes (using the Richter scale), determining the pH of a solution in chemistry, and modeling exponential decay in physics. The general form of a logarithmic equation can be expressed as $\log_b(f(x)) = g(x)$, where $b$ is the base of the logarithm, $f(x)$ is an expression involving the variable $x$, and $g(x)$ can also be an expression involving $x$ or a constant. The key to solving these equations lies in our ability to manipulate logarithms and use their properties to isolate the variable.
Logarithms are the inverse functions of exponentials. The expression $\log_b(a) = c$ is equivalent to $b^c = a$. This inverse relationship is the cornerstone of solving logarithmic equations. The base $b$ must be a positive number not equal to 1, and $a$ must be positive, because logarithms are only defined for positive arguments. When we encounter an equation involving logarithms, we often try to use properties of logarithms to simplify it. For instance, if we have $\log_b(x) = \log_b(y)$, and both $x$ and $y$ are positive, then we can conclude that $x = y$. This property is especially useful when we have logarithms with the same base on both sides of the equation.
However, solving logarithmic equations is not merely a matter of applying formulas. We must also be careful about the domain of logarithmic functions. The argument of a logarithm must be positive. Thus, if our equation involves terms like $\log(f(x))$, we must ensure that $f(x) > 0$. This requirement means that after solving the equation, we need to check our solutions to make sure they do not lead to taking the logarithm of a non-positive number. Solutions that violate this condition are called extraneous solutions and must be discarded. Therefore, a complete solution to a logarithmic equation involves not only finding potential values for the variable but also verifying that these values satisfy the domain restrictions imposed by the logarithms.
Solving the Equation $\log _5(10 x-1)=\log _5(9 x+7)$
Now, let's tackle the given equation: $\log _5(10 x-1)=\log _5(9 x+7)$. This equation involves logarithms with the same base (base 5) on both sides, which makes it a prime candidate for using the property that if $\log_b x = \log_b y$, then $x = y$, provided $x$ and $y$ are positive. Applying this property will allow us to eliminate the logarithms and transform the equation into a simpler algebraic form.
Step 1: Apply the Property of Logarithms
Since the bases of the logarithms are the same, we can equate the arguments of the logarithms: $10x - 1 = 9x + 7$. This step is justified because the logarithmic function is one-to-one, meaning that if the logarithms of two expressions are equal, then the expressions themselves must be equal. However, we must remember that this step is valid only if both $10x - 1$ and $9x + 7$ are positive. We will verify this condition later when we check our solution.
Step 2: Solve the Resulting Algebraic Equation
Now we have a linear equation in terms of $x$. Let's solve for $x$:
Subtract $9x$ from both sides:
Add 1 to both sides:
So, we have found a potential solution: $x = 8$. However, as mentioned earlier, we must check this solution to ensure it does not lead to taking the logarithm of a non-positive number.
Step 3: Check the Solution
To check our solution, we substitute $x = 8$ back into the original logarithmic equation and verify that the arguments of the logarithms are positive:
For the left side, $10x - 1 = 10(8) - 1 = 80 - 1 = 79$, which is positive.
For the right side, $9x + 7 = 9(8) + 7 = 72 + 7 = 79$, which is also positive.
Since both arguments are positive when $x = 8$, our solution is valid. We can also substitute $x=8$ into the original equation to confirm:
Both sides are equal, so $x = 8$ is indeed the solution to the equation.
Detailed Explanation of Each Step
To ensure a comprehensive understanding, let's revisit each step with a more detailed explanation:
Step 1: Apply the Property of Logarithms
This step is crucial because it allows us to transform the logarithmic equation into an algebraic equation. The property we used states that if $\log_b x = \log_b y$, then $x = y$. This property is a direct consequence of the fact that the logarithmic function is one-to-one. A one-to-one function is a function that never takes the same value twice. In other words, if $f(x_1) = f(x_2)$, then $x_1 = x_2$. The logarithmic function is one-to-one because its inverse, the exponential function, is one-to-one. When we apply this property, we are essentially undoing the logarithmic operation, which simplifies the equation significantly.
However, it is essential to remember the conditions under which this property holds. The arguments $x$ and $y$ must be positive, and the base $b$ must be a positive number not equal to 1. These conditions arise from the definition of the logarithm. The logarithm $\log_b a$ is defined as the exponent to which we must raise $b$ to obtain $a$. Since exponential functions with a positive base can only produce positive results, the argument $a$ must be positive. Furthermore, the base $b$ cannot be 1 because $1$ raised to any power is still $1$, which would make the logarithm undefined for all numbers other than 1. The base $b$ must also be positive to avoid issues with non-real results.
Step 2: Solve the Resulting Algebraic Equation
After applying the logarithmic property, we are left with a linear equation. Solving a linear equation involves isolating the variable on one side of the equation. We do this by performing the same operations on both sides of the equation, maintaining the equality. In this case, we first subtracted $9x$ from both sides to collect the terms involving $x$ on one side. Then, we added $1$ to both sides to isolate the term with $x$. These steps are standard algebraic manipulations that are used to solve a wide variety of equations.
The goal of these manipulations is to get the equation into the form $x = c$, where $c$ is a constant. This form gives us the value of the variable directly. In our case, after these steps, we found that $x = 8$. This is a potential solution, but it is crucial to remember that we have not yet verified whether this solution is valid in the original logarithmic equation.
Step 3: Check the Solution
Checking the solution is a critical step in solving logarithmic equations. This is because logarithmic functions have a restricted domain. The argument of a logarithm must be positive. If we plug in a value for $x$ that makes the argument of any logarithm in the equation non-positive, then that value is not a valid solution. Such solutions are called extraneous solutions.
To check the solution, we substitute the value we found for $x$ back into the original equation and verify that the arguments of all logarithms are positive. In our case, we found that when $x = 8$, both $10x - 1$ and $9x + 7$ are equal to $79$, which is positive. This means that $x = 8$ is a valid solution. Additionally, we substituted $x=8$ back into the original equation to confirm that both sides of the equation are equal. This final verification step provides added assurance that our solution is correct.
Conclusion
In conclusion, the solution to the equation $\log _5(10 x-1)=\log _5(9 x+7)$ is $x = 8$. We arrived at this solution by first applying the property of logarithms that states if $\log_b x = \log_b y$, then $x = y$, provided $x$ and $y$ are positive. This allowed us to transform the logarithmic equation into a linear equation, which we then solved for $x$. Finally, we checked our solution to ensure that it did not lead to taking the logarithm of a non-positive number.
Solving logarithmic equations requires a careful understanding of the properties of logarithms and the domain restrictions they impose. By following a systematic approach, including checking solutions, we can accurately solve these types of equations. The methods and principles discussed in this article provide a solid foundation for tackling more complex logarithmic equations and problems in various fields of mathematics and science.
The ability to solve logarithmic equations is a valuable skill that has applications in many areas. Whether you are working on mathematical problems, scientific research, or engineering designs, understanding how to manipulate and solve logarithmic equations will prove to be a valuable asset. Remember to always check your solutions and pay attention to the domain restrictions to ensure the validity of your results.
