/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Express the equation in logarith... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Express the equation in logarithmic form. (a) \(e^{x+1}=0.5\) (b) \(e^{0.5 x}=t\)

Short Answer

Expert verified
(a) \(x = \ln(0.5) - 1\), (b) \(x = \frac{\ln(t)}{0.5}\).

Step by step solution

01

Identify the Exponential Equation

The given equations are exponential equations with the base as the natural number \(e\).
02

Convert to Logarithmic Form (a)

For the equation \(e^{x+1} = 0.5\), apply the definition of logarithms. The formula is \(a^b = c\) can be rewritten as \(b = \log_a(c)\). Therefore, \(e^{x+1} = 0.5\) can be rewritten as \(x + 1 = \ln(0.5)\).
03

Solve for x (a)

Isolate \(x\) by subtracting 1 from both sides. Hence, \(x = \ln(0.5) - 1\).
04

Convert to Logarithmic Form (b)

For the equation \(e^{0.5x} = t\), again apply the definition of logarithms. Therefore, the equation is rewritten as \(0.5x = \ln(t)\).
05

Solve for x (b)

Isolate \(x\) by dividing both sides by 0.5. Thus, \(x = \frac{\ln(t)}{0.5}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Exponential Equations
Exponential equations are equations where a constant base is raised to a variable exponent. These equations often take the general form \(a^b = c\), where \(a\) is a constant, \(b\) is a variable, and \(c\) is a result. In many mathematical scenarios, especially in scientific contexts, the base \(e\), approximately equal to 2.718, is used. This is known as the natural exponential base. Exponential equations with this base are referred to as natural exponential equations. Examples include:
  • \(e^{x} = 3\)
  • \(e^{2x + 1} = 10\)
These equations often arise in growth processes like population growth or radioactive decay, where the change is proportional to the amount present. Solving exponential equations typically involves converting them into a logarithmic form. This transformation is a crucial step in finding the unknown variable effectively.
Natural Logarithm
A natural logarithm is a logarithm with base \(e\). It is denoted as \(\ln(x)\) rather than \(\log_e(x)\), offering a convenient shorthand. The natural logarithm is widely used due to its properties related to exponential functions and calculus, simplifying the process of differentiating or integrating exponential equations. The natural logarithm answers the question: "To what power must \(e\) be raised to yield \(x\)?" For example, \(\ln(1) = 0\) because \(e^0 = 1\), and \(\ln(e) = 1\) because \(e^1 = e\). Properties of the natural logarithm include:
  • \(\ln(ab) = \ln(a) + \ln(b)\)
  • \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
  • \(\ln(a^n) = n\ln(a)\)
These rules are particularly valuable in simplifying complex logarithmic expressions.
Converting to Logarithmic Form
Converting an exponential equation to its logarithmic form allows for easier manipulation and solution of the variable exponent. The conversion is based on the principle that any exponential equation \(a^b = c\) can be rewritten as \(b = \log_a(c)\). For natural exponential equations where the base is \(e\), this becomes \(b = \ln(c)\). ### Conversion Steps: 1. **Identify the Exponential Equation:** Recognize the base \(e\) and the structure \(e^{b} = c\). 2. **Apply the Logarithmic Definition:** Rewrite the exponent in terms of its logarithm using \(b = \ln(c)\). Once in logarithmic form, solving for the variable often involves basic algebraic manipulations, making it clear and easy to isolate the variable. This technique is especially helpful in cases where direct calculation through exponents can be cumbersome or impractical. By converting to a logarithmic form, the process of solving for unknowns becomes not only simpler but also more intuitive.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Compound Interest If \(\$ 600\) is invested at an interest rate of \(2.5 \%\) per year, find the amount of the investment at the end of 10 years for the following compounding methods. (a) Annually (b) Semiannually (c) Quarterly (d) Continuously

These exercises use Newton’s Law of Cooling. Newton's Law of Cooling is used in homicide investigations to determine the time of death. The normal body temperature is \(98.6^{\circ} \mathrm{F}\). Immediately following death, the body begins to cool. It has been determined experimentally that the constant in Newton's Law of Cooling is approximately \(k=0.1947,\) assuming that time is measured in hours. Suppose that the temperature of the surroundings is \(60^{\circ} \mathrm{F}\). (a) Find a function \(T(t)\) that models the temperature \(t\) hours after death. (b) If the temperature of the body is now \(72^{\circ} \mathrm{F}\), how long ago was the time of death?

A certain strain of bacteria divides every three hours. If a colony is started with 50 bacteria, then the time \(t\) (in hours) required for the colony to grow to \(N\) bacteria is given by $$t=3 \frac{\log (N / 50)}{\log 2}$$ Find the time required for the colony to grow to a million bacteria.

Atmospheric Pressure Atmospheric pressure \(P\) (in kilopascals, kPa) at altitude \(h\) (in kilometers, \(\mathrm{km}\) ) is governed by the formula $$\ln \left(\frac{P}{P_{0}}\right)=-\frac{h}{k}$$ where \(k=7\) and \(P_{0}=100 \mathrm{kPa}\) are constants. (a) Solve the equation for \(P\). (b) Use part (a) to find the pressure \(P\) at an altitude of \(4 \mathrm{km}\)

These exercises deal with logarithmic scales. The pH readings for wines vary from 2.8 to 3.8. Find the corresponding range of hydrogen ion concentrations.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.