Problem 26
Question
In \(21-32,\) for each given logarithm, find \(x,\) the antilogarithm. Write the answer to four decimal places. $$ \ln x=2.5619 $$
Step-by-Step Solution
Verified Answer
The antilogarithm is approximately 12.9507.
1Step 1: Understanding Natural Logarithm
The given equation is \( \ln x = 2.5619 \). Here, \( \ln x \) is the natural logarithm of \( x \). The base of the natural logarithm is \( e \), where \( e \approx 2.71828 \). Finding the antilogarithm involves reversing the logarithmic function.
2Step 2: Exchanging Logarithm with Exponential Form
Convert the logarithmic equation into its exponential form: \( e^{2.5619} = x \). The exponential form is derived directly from the property of logarithms: \( \ln x = y \) implies \( x = e^y \).
3Step 3: Calculating the Antilogarithm
Using the exponential form \( x = e^{2.5619} \), calculate the value using a calculator or computational tool. Evaluate \( e^{2.5619} \) to obtain a numerical value for \( x \).
4Step 4: Final Calculation and Rounding
After computing the expression \( e^{2.5619} \), you get approximately \( 12.9507 \). Make sure to round this to four decimal places as required, thus \( x = 12.9507 \).
Key Concepts
AntilogarithmExponential FormE ConstantLogarithmic Properties
Antilogarithm
When we talk about antilogarithms, we're discussing the inverse operation of a logarithm. If you have \(\ln x = 2.5619\), the task is to find the number \(x\) such that, when you apply the natural logarithm to it, you'll get 2.5619. In simpler terms, determining the antilogarithm is like "undoing" the logarithm in order to retrieve the original value, \(x\). This process is crucial because it helps translate logarithmic equations back into their more familiar numerical forms. Understanding this concept allows you to solve problems where logarithm values are involved, by reversing that process to find the original number.
Exponential Form
The exponential form is a way of expressing numbers, where the power or exponent is shown explicitly. Given the logarithmic equation \(\ln x = 2.5619\), you can convert it into the exponential form \(e^{2.5619} = x\). Think of this step as moving from a logarithmic perspective back to a numerical value, where \(x\) is calculated by raising the constant \(e\) to the given power. This transformation is based on the fundamental relationship between exponents and logarithms: \(\log_b(a) = c\) implies \(b^c = a\). So, when working with natural logarithms, you'll see \(e^y = x\) when switching to exponential form.
E Constant
The \(e\) constant is one of mathematics' most intriguing numbers. Approximately equal to 2.71828, \(e\) is the base of the natural logarithm. It frequently appears in growth and decay problems, as well as in many mathematical and scientific contexts. It is known as Euler's number, and its properties make it uniquely suited to dealing with exponential growth calculations. The constant \(e\) is irrational, meaning it cannot be expressed as a simple fraction, and its decimal expansion goes on forever without repeating. When calculating the antilogarithm using \(e\), you're essentially using this constant in action.
Logarithmic Properties
Logarithmic properties are fundamental rules used to manipulate logarithmic expressions. These rules make it easier to solve equations and derive results such as exponential forms. Here are key properties:
- Product Property: \(\log_b(MN) = \log_b(M) + \log_b(N)\)
- Quotient Property: \(\log_b(M/N) = \log_b(M) - \log_b(N)\)
- Power Property: \(\log_b(M^k) = k \cdot \log_b(M)\)
Other exercises in this chapter
Problem 25
In \(15-26,\) write each logarithmic equation in exponential form. $$ \log _{49} 343=\frac{3}{2} $$
View solution Problem 25
Write each expression as a single logarithm. \(\log _{2} a+\log _{2} b\)
View solution Problem 26
In \(24-35,\) for each given logarithm, find the antilogarithm, \(x .\) Write the answer to four decimal places. $$ \log x=1.3826 $$
View solution Problem 26
In \(15-26,\) write each logarithmic equation in exponential form. $$ -\frac{2}{5}=\log _{32} 0.25 $$
View solution