Problem 62 - College Algebra and Calculus: An Applied Approach

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln x=5\)

Step-by-Step Solution

Answer

Therefore, the solution to the equation is \(x \approx 148.413\).

Step 1: Understand the equation

The equation provided is in logarithmic form. The logarithm in use is the natural logarithm which has the base of euler's number \(e\). The equation to solve is \(\ln x = 5\). The goal is to express \(x\) in terms of \(e\).

Step 2: Convert into exponential form

The equation \(\ln x = 5\) can be rewritten in exponential form by taking the base of the natural logarithm \(e\) and making it the base of an exponent. This gives us the equation \(e^5 = x\).

Step 3: Solve for x

Having the equation \(e^5 = x\), solving for \(x\) simply involves evaluating \(e^5\). Remember to approximate to three decimal places as the problem instructs.

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \(\ln x\), is a mathematical function that uses the constant \(e\), approximately equal to 2.71828, as its base. This unique number, Euler's number, is significant in mathematics due to its natural occurrence in various growth processes, from finance to physics. In the context of solving logarithmic equations, understanding the natural logarithm is crucial because it tells you how many times you must multiply \(e\) by itself to get the number \(x\).

While working with \(\ln\), remember that \(\ln e = 1\) because \(e\) raised to the power of 1 is \(e\). Similarly, \(\ln 1 = 0\) because any number raised to the power of 0 is 1, and \(e^0 = 1\). In the exercise \(\ln x = 5\), the base \(e\) is implied and does not need to be written out, which is a unique characteristic of natural logarithms.

Exponential Form

Converting a logarithmic equation to exponential form is a valuable technique when solving logarithmic equations. The exponential form is expressed as \(b^y = x\), where \(b\) is the base, \(y\) is the exponent, and \(x\) is the result of \(b\) raised to the power of \(y\). When dealing with natural logarithms, \(e\) is always the base because \(\ln x\) refers to the power to which \(e\) must be raised to obtain \(x\).

So, when you have an equation like \(\ln x = 5\), you can translate this into \(e^5 = x\). Here, \(5\) is the exponent to which \(e\) is raised to get \(x\). This step is crucial in the solution process as it sets up the equation in a form where \(x\) can be isolated and calculated directly.

Algebraic Manipulation

Algebraic manipulation involves using mathematical operations to rearrange equations and solve for unknown variables. This process can include adding, subtracting, multiplying, dividing, factoring, expanding, or using properties of exponents and logarithms to isolate the variable of interest.

In our exercise \(\ln x = 5\), converting to exponential form is part of algebraic manipulation. After converting, solving for \(x\) becomes more manageable as you're dealing with a clearer and more direct equation, \(e^5 = x\). Here, no further algebraic manipulation is needed, and you would simply calculate \(e^5\) to find the value of \(x\). However, in more complex logarithmic equations, additional algebraic steps might be necessary to solve for the variable.