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Chapter 4: Problem 110

Solve each equation. Find the exact solutions. $$\ln \left(\frac{1}{\sqrt{e}}\right)=x$$

Short Answer

Expert verified
x = -1/2

Step by step solution

01

Understand the equation

The given equation is \( \ln \( \frac{1}{\sqrt{e}}\) = x\). The goal is to solve for x by using logarithmic and algebraic properties.
02

Simplify the argument

Recognize that \( \frac{1}{\sqrt{e}}\) can be rewritten using exponents. Recall that \( \sqrt{e} = e^{1/2}\) and therefore \( \frac{1}{\sqrt{e}} = \frac{1}{e^{1/2}}\) which simplifies to \( e^{-1/2}\).
03

Apply the logarithm property

Apply the natural logarithm to \( e^{-1/2}\). Using the property that \( \ln \( e^y \) = y\), we find that \( \ln \left(e^{-1/2}\right) = -1/2\).
04

State the solution

Since \( \ln \left(e^{-1/2}\right) = -1/2\), the solution to the equation is \( x = -1/2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
In mathematics, the natural logarithm, denoted as \(\text{ln}\), is a specific logarithm that has the constant \(e\) (approximately 2.71828) as its base. This function is especially useful in calculus and higher-level math involving growth and decay processes. For example, when you see \(\text{ln}(e^y)\), it simply means the natural logarithm of \(e\) raised to some power \(y\). A critical property to remember is:
  • \(\text{ln}(e^y) = y\)
This property stems from the definition of logarithms and the fact that \(\text{ln}\) is the inverse function of the exponential function with base \(e\). In other words, it
Exponential Expressions
Exponential expressions involve numbers or variables raised to a power. In our example, we deal with \(e\) raised to various powers. Here's how you manipulate these expressions: Understanding that \(\frac{1}{\text{something}}\) is equivalent to raising that something to the power of -1 is crucial. Additionally, understanding fractional exponents is vital. For example:
  • \(\frac{1}{\text{something}}\) = \(\text{something}^{-1}\)
  • \(\frac{1}{\text{something}^{a}}\) = \(\text{something}^{-a}\)
By recognizing the above, you can rewrite \(\frac{1}{\text{鈭歟}}\) as \(e^{-1/2}\). Such manipulation makes it easier to apply logarithmic properties later in the problem.
Logarithm Properties
Logarithms have several key properties that are often used to simplify and solve equations. Key properties include:
  • \(\text{ln}(a \times b) = \text{ln}(a) + \text{ln}(b)\)
  • \(\text{ln}(a / b) = \text{ln}(a) - \text{ln}(b)\)
  • \(\text{ln}(a^b) = b \times \text{ln}(a)\)
In our specific case, we used the property \(\text{ln}(e^y) = y\). This property holds because ln(x) is defined as the inverse function of the exponential function \(e^x\). When simplifying the equation \(\text{ln}\bigg(\frac{1}{\text{鈭歟}}\bigg)\), we first rewrote \(\frac{1}{\text{鈭歟}}\) as \(e^{-1/2}\). Then, applying \(\text{ln}\), we used \(\text{ln}(e^{-1/2}) = -1/2\). Understanding these properties is crucial for efficiently solving logarithmic equations.

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Most popular questions from this chapter

Solve each problem. When needed, use 365 days per year and 30 days per month. Present Value Compounding Daily A credit union pays \(6.5 \%\) annual interest compounded daily. What deposit today (present value) would amount to \(\$ 3000\) in 5 years and 4 months? Hint Solve the compound interest formula for \(P\).

Solve each equation. Find the exact solutions. $$\log _{2}\left(\log _{2}\left(\log _{2}\left(2^{\left(4^{\circ}\right)}\right)\right)\right)=3$$

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