Solving Logarithmic Equations With One-to-One Property Example

This article delves into the method of solving logarithmic equations using the one-to-one property of logarithms. We will specifically address the equation $\ln(n+18) - \ln(n) = \ln(16)$, demonstrating a step-by-step approach to finding the solution. Understanding the properties of logarithms is crucial for solving such equations efficiently. This method allows us to simplify complex logarithmic expressions and isolate the variable, ultimately leading to the solution.

Understanding Logarithmic Properties

Before we dive into solving the equation, let's briefly review some essential logarithmic properties. These properties are the foundation for manipulating and simplifying logarithmic expressions. The key property we'll use in this example is the quotient rule of logarithms, which states that the logarithm of a quotient is equal to the difference of the logarithms. Mathematically, this is expressed as $\log_b(x) - \log_b(y) = \log_b(\frac{x}{y})$. Another crucial property is the one-to-one property, which dictates that if $\log_b(x) = \log_b(y)$, then $x = y$. This property allows us to eliminate logarithms from an equation when we have a single logarithm on each side with the same base.

In addition to the quotient rule and the one-to-one property, it's also helpful to remember the definition of a logarithm. The expression $\log_b(x) = y$ is equivalent to $b^y = x$, where $b$ is the base of the logarithm. Understanding this relationship helps us convert between logarithmic and exponential forms, which can be useful in certain situations. Furthermore, the power rule of logarithms, $\log_b(x^p) = p\log_b(x)$, and the product rule of logarithms, $\log_b(xy) = \log_b(x) + \log_b(y)$, are valuable tools for manipulating logarithmic expressions. By mastering these properties, you'll be well-equipped to tackle a wide range of logarithmic equations.

The natural logarithm, denoted by $\ln$, is a logarithm with base $e$, where $e$ is an irrational number approximately equal to 2.71828. The properties of natural logarithms are the same as those for logarithms with any other base. Therefore, the quotient rule, one-to-one property, power rule, and product rule all apply to natural logarithms as well. The ability to work comfortably with natural logarithms is essential, as they appear frequently in calculus, physics, and other areas of mathematics and science. Familiarity with these properties will not only help you solve equations but also deepen your understanding of logarithmic functions and their applications.

Solving the Equation: $\ln(n+18) - \ln(n) = \ln(16)$

Now, let's apply our knowledge of logarithmic properties to solve the given equation: $\ln(n+18) - \ln(n) = \ln(16)$. Our primary goal is to isolate the variable n. The first step involves using the quotient rule of logarithms to combine the left side of the equation. Recall that $\ln(a) - \ln(b) = \ln(\frac{a}{b})$. Applying this rule to our equation, we get:

$\ln(\frac{n+18}{n}) = \ln(16)$

Now that we have a single natural logarithm on both sides of the equation, we can utilize the one-to-one property of logarithms. This property states that if $\ln(x) = \ln(y)$, then $x = y$. Therefore, we can remove the natural logarithms from both sides of the equation:

$\frac{n+18}{n} = 16$

Our next step is to solve this algebraic equation for n. We begin by multiplying both sides of the equation by n to eliminate the fraction:

$n + 18 = 16n$

Now, we need to isolate the terms containing n on one side of the equation. Subtracting n from both sides, we have:

$18 = 15n$

Finally, to solve for n, we divide both sides of the equation by 15:

$n = \frac{18}{15}$

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3:

$n = \frac{6}{5}$

Therefore, the solution to the equation $\ln(n+18) - \ln(n) = \ln(16)$ is $n = \frac{6}{5}$, which can also be expressed as the decimal 1.2.

Verification and Domain Considerations

It's crucial to verify our solution by plugging it back into the original equation. This step ensures that our solution is valid and doesn't introduce any extraneous solutions. Substituting $n = \frac{6}{5}$ into the original equation, we get:

$\ln(\frac{6}{5} + 18) - \ln(\frac{6}{5}) = \ln(16)$

First, we simplify the expression inside the first logarithm:

$\frac{6}{5} + 18 = \frac{6}{5} + \frac{90}{5} = \frac{96}{5}$

So, the equation becomes:

$\ln(\frac{96}{5}) - \ln(\frac{6}{5}) = \ln(16)$

Using the quotient rule of logarithms again, we have:

$\ln(\frac{\frac{96}{5}}{\frac{6}{5}}) = \ln(16)$

Simplifying the fraction inside the logarithm:

$\ln(\frac{96}{5} \cdot \frac{5}{6}) = \ln(16)$

$\ln(16) = \ln(16)$

Since this equation holds true, our solution $n = \frac{6}{5}$ is verified.

In addition to verification, it's also important to consider the domain of the logarithmic functions in the equation. Logarithms are only defined for positive arguments. Therefore, we must ensure that $n + 18 > 0$ and $n > 0$. Our solution, $n = \frac{6}{5}$, satisfies both of these conditions. If we had obtained a negative solution or a solution that made either $n + 18$ or $n$ non-positive, we would have had to discard it as an extraneous solution.

Conclusion

In this article, we successfully solved the logarithmic equation $\ln(n+18) - \ln(n) = \ln(16)$ using the one-to-one property of logarithms. The steps involved applying the quotient rule to combine the logarithms on the left side, using the one-to-one property to eliminate the logarithms, solving the resulting algebraic equation, and verifying the solution. We also emphasized the importance of considering the domain of logarithmic functions to avoid extraneous solutions. The solution to the equation is $n = \frac{6}{5}$ or 1.2. By understanding and applying the properties of logarithms, you can effectively solve a wide range of logarithmic equations. Remember to always verify your solutions and consider the domain restrictions to ensure accuracy.

This method provides a structured approach to tackling logarithmic equations, making them less intimidating and more manageable. The key takeaway is the power of logarithmic properties in simplifying complex expressions and isolating the variable. With practice and a solid grasp of these properties, you can confidently navigate logarithmic equations in various mathematical contexts.