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Chapter 3: Problem 84

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\ln x+1=0$$

Short Answer

Expert verified
The approximate solution to the equation \( \ln x + 1 = 0 \) is \( x = 0.368 \)

Step by step solution

01

Isolate Natural Logarithm

The first step is to isolate the natural logarithm on one side of the equation. To do this, deduct 1 from both sides of the given equation \( \ln x + 1 = 0 \). After this, the equation transforms to \( \ln x = -1 \)
02

Convert to Exponential Function

The next step is to transition from logarithmic form to exponential form. This is done using the natural exponent \( e \). This means \( e^{-1} = x \)
03

Calculate the Value of x

Now the equation can be evaluated numerically. The approximation to three decimal places of \( e^{-1} \) is 0.368, so \( x = 0.368 \)

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Key Concepts

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

Natural Logarithms
Natural logarithms are a special type of logarithm that uses the constant \( e \) (approximately equal to 2.718) as the base. The natural logarithm of a number \( x \) is expressed as \( \ln x \). These logarithms are used frequently in calculus and real-world applications like exponential growth or decay.

Here are some properties of natural logarithms to keep in mind:
  • \( \ln 1 = 0 \) because \( e^0 = 1 \).
  • \( \ln e = 1 \) because \( e^1 = e \).
  • \( \ln(ab) = \ln a + \ln b \) - this property simplifies multiplication inside a logarithm.
  • \( \ln(a^b) = b \ln a \) - an exponent can be brought in front as a multiplier.

In the problem \( \ln x + 1 = 0 \), isolating \( \ln x \) lets us directly solve for \( x \) using these properties.
Exponential Functions
Exponential functions involve variables as exponents, usually resulting in rapid increases or decreases.

The base \( e \) is used frequently in these functions due to its natural properties that make calculus calculations smoother. In the equation \( e^{-1} = x \), we switch from logarithmic form \( \ln x = -1 \) to exponential form to find \( x \).

Exponential functions are significant because:
  • They model growth and decay, such as population growth or radioactive decay.
  • \( e^x \) is its own derivative and integral, making it unique.
  • The inverse of an exponential function is a logarithm, which helps in solving equations like the one at hand.

Mastering the conversion between logarithmic and exponential forms is essential in solving and understanding equations involving natural logarithms.
Algebraic Solutions
Algebraic solutions are all about step-by-step manipulation of equations to find the value of a variable. This process uses the rules of algebra and the properties of the functions involved.

When solving \( \ln x + 1 = 0 \), the key algebraic steps are:
  • Isolate the logarithmic expression, leading to \( \ln x = -1 \).
  • Convert the equation to exponential form to solve for \( x \), resulting in \( x = e^{-1} \).
  • Calculate \( e^{-1} \) to find the approximate decimal value, which is 0.368.
These steps reflect the essence of algebraic solutions: breaking down complex equations into simpler, solvable steps.

Algebraic solutions are vital because they offer a methodical way to handle and solve different types of equations using consistent rules.

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Most popular questions from this chapter

A $$\$ 120,000$$ home mortgage for 30 years at \(7 \frac{1}{2} \%\) has a monthly payment of $$\$ 839.06 .$$ Part of the monthly payment is paid toward the interest charge on the unpaid balance, and the remainder of the payment is used to reduce the principal. The amount that is paid toward the interest is \(u=M-\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}\) and the amount that is paid toward the reduction of the principal is \(v=\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}\) In these formulas, \(P\) is the size of the mortgage, \(r\) is the interest rate, \(M\) is the monthly payment, and \(t\) is the time (in years). (a) Use a graphing utility to graph each function in the same viewing window. (The viewing window should show all 30 years of mortgage payments.) (b) In the early years of the mortgage, is the larger part of the monthly payment paid toward the interest or the principal? Approximate the time when the monthly payment is evenly divided between interest and principal reduction. (c) Repeat parts (a) and (b) for a repayment period of 20 years \((M=\$ 966.71) .\) What can you conclude?

Use the Richter scale \(R=\log \frac{l}{I_{0}}\) for measuring the magnitudes of earthquakes. Find the magnitude \(R\) of each earthquake of intensity \(I\) (let \(I_{0}=1\) ). (a) \(I=199,500,000\) (b) \(I=48,275,000\) (c) \(I=17,000\)

Use your school's library, the Internet, or some other reference source to write a paper describing John Napier's work with logarithms.

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$6 \log _{3}(0.5 x)=11$$

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$\ln (x+1)=2-\ln x$$
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