Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln x-\ln (x+2)=3\)

Short answer

The solution to the equation is approximately x = 12.577

Step-by-step solution

Combine the logarithms

The given equation is \( \ln x-\ln (x+2)=3 \). Using the logarithm rule \( \ln a - \ln b = \ln \frac{a}{b} \), the equation is transformed to \( \ln \frac{x}{x+2} = 3 \)

Use rules of logarithms

Next, we rewrite the logarithm into an exponential form. This is done using the rule \( \ln a = b \) which can be rewritten as \( e^b = a \). Hence, the equation \( \ln \frac{x}{x + 2} = 3 \) becomes \( e^3 = \frac{x}{x+2} \)

Isolate variable x

To isolate x, we multiply both sides by \( x+2 \). We have: \( e^3 * (x+2) = x \). That is, \( e^3*x + 2*e^3 = x \). Re-arranging this gives us \( e^3*x -x = 2*e^3 \). Simplify by collecting like terms together: \( x*(e^3 - 1) = 2*e^3 \)

Find the approximate value for x

Finally, we find x by dividing both sides by \( e^3 - 1 \): \( x = \frac{2*e^3}{e^3 - 1} \). Using a calculator, this evaluates to approximately 12.577

Key concepts

Logarithmic and Exponential Functions

Understanding the interplay between logarithmic and exponential functions is essential for solving equations that involve logarithms. Considering our equation \(\ln x - \ln (x+2) = 3\) as an example, it's important to recognize that the natural logarithm, denoted \ln, and the base \(e\) exponential function are inverse operations. What this means is that for any positive number \(a\), if we have \(\ln a = b\), we can also write this as \(e^b = a\).

This concept turns the process of solving a logarithmic equation into finding the exponent that the base \(e\) must be raised to in order to obtain the argument of the logarithm. It also means that logarithmic functions can be used to 'undo' the effect of an exponential function, particularly useful in isolating variables as seen in later steps of solving the given exercise.

Algebraic Manipulation

Algebraic manipulation encompasses various techniques used to simplify or rearrange equations and expressions to solve for unknown variables. This includes operations like combining like terms, factoring, expanding, and using arithmetic properties. In the context of our logarithmic equation, we start out by using the property of logarithms that allows us to combine two logs with the same base into a single log by dividing the arguments, obtaining \(\ln \frac{x}{x+2} = 3\).

This step showcases how algebraic manipulation can make complex expressions more manageable and set the stage for using different mathematical properties, such as those governing logarithmic and exponential relationships, to further solve for the variable.

Properties of Logarithms

The properties of logarithms are powerful tools that make it easier to work with logarithmic expressions. These properties, rooted in the fundamental characteristics of logarithms, include the product, quotient, and power rules. In the exercise above, the quotient rule was applied in step 1 to combine the logs. The rule states, for any positive numbers \(a\) and \(b\) and base \(c\), that \(\log_c\frac{a}{b} = \log_ca - \log_cb\).

By utilizing these properties correctly, logarithms can be simplified or expanded as necessary to solve an equation. Step 2 of our solution transforms the logarithmic equation into an exponential one, illustrating how the fundamental definition of logarithms can bridge the gap between different types of functions for the purpose of finding solutions to variables.

Isolation of Variables

The isolation of variables is a cornerstone technique in solving equations. It involves manipulating the equation to get the unknown variable on one side, allowing us to calculate its value. For the given problem, after transforming the logarithmic equation into an exponential one, we multiplied both sides by \(x+2\) to move toward isolating \(x\).

This process meant dealing with an expression where the variable appeared in more than one term, leading to the necessity of gathering like terms together and factoring out the variable. Eventually, as shown in Step 4, we divide by the coefficient of \(x\), here being \(e^3 - 1\), to isolate \(x\). This is one of the many scenarios in algebra where isolating the variable can require multiple steps and a good grasp of algebraic manipulation as well as properties of special functions like logarithms.