/*! 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 36 Exer. \(21-36:\) Solve the equat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Exer. \(21-36:\) Solve the equation. $$e^{x \ln 2}=0.25$$

Short Answer

Expert verified
The solution is \( x = -2 \).

Step by step solution

01

Understand the Given Equation

The given equation is \( e^{x \ln 2} = 0.25 \). Our goal is to solve for \( x \). This involves understanding that the exponent in the expression is using the natural base \( e \) and logarithmic properties.
02

Simplify the Equation Using Logarithms

To solve for \( x \), take the natural logarithm on both sides of the equation: \( \ln(e^{x \ln 2}) = \ln(0.25) \). Using the property of logarithms that \( \ln(e^a) = a \), we simplify the left side to be \( x \ln 2 \). Thus, the equation becomes: \( x \ln 2 = \ln(0.25) \).
03

Solve for x

Now, we solve for \( x \) by dividing both sides of the equation by \( \ln 2 \). So, \( x = \frac{\ln(0.25)}{\ln 2} \).
04

Calculate the Solution

Calculate \( \ln(0.25) \) and \( \ln 2 \) using a calculator. We find that \( \ln(0.25) = -1.3863 \) and \( \ln 2 = 0.6931 \). Therefore, \( x = \frac{-1.3863}{0.6931} \approx -2 \).

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.

Natural Logarithm
The natural logarithm, denoted as \(\ln\), is a function that reverses the exponential function with base \(e\). The constant \(e\) is an irrational number approximately equal to 2.71828. Essentially, the natural logarithm tells us what power we need to raise \(e\) to get a certain number.
\[ \ln(e^a) = a \]
When we take the natural logarithm of an exponential expression with base \(e\), such as \(e^a\), it simplifies to \(a\). This is a key property that helps in solving exponential equations as it linearizes the expression, making it easier to isolate the variable involved.
Logarithmic Properties
Logarithmic properties are essential tools in algebra that help simplify complex expressions. Here are some key properties relevant to solving exponential equations:
  • \( \ln(AB) = \ln A + \ln B \) - The logarithm of a product is the sum of the logarithms.
  • \( \ln\left(\frac{A}{B}\right) = \ln A - \ln B \) - The logarithm of a quotient is the difference of the logarithms.
  • \( \ln(A^b) = b \ln A \) - The logarithm of a power is the exponent times the logarithm of the base.
By applying these properties, particularly simplifying a power expression such as \(e^{x \ln 2}\), it becomes manageable to solve complex equations by reducing exponential expressions to linear ones.
Exponential Functions
Exponential functions are mathematical expressions involving constants raised to a variable power. A common form is \(a^x\), where \(a\) is a constant. These functions model various phenomena such as population growth, radioactive decay, and compound interest.
The key feature of exponential functions is their rate of growth. For the natural exponential function \(e^x\), it grows rapidly as \(x\) increases. When graphed, it forms a curve that becomes increasingly steep.
In our exercise, acknowledging the function's properties allowed us to align the equation \(e^{x \ln 2} = 0.25\) with logarithmic principles to solve for the unknown variable.
Solving Equations
Solving exponential equations typically involves reducing them into linear forms where the variable becomes more accessible. The process often starts with isolating the exponential term then applying the natural logarithm to both sides:
1. If you have an equation like \(e^{x \ln 2} = 0.25\), take \(\ln\) on both sides to use the logarithm properties.
2. Simplifying using \(\ln(e^a) = a\) gives us \(x \ln 2\).
3. Isolate \(x\) by dividing each side by \(\ln 2\), leading to \(x = \frac{\ln(0.25)}{\ln 2}\).
By decomposing the equation step-by-step, we efficiently calculate \(x\) without computational strain. This technique illustrates the strength of logarithms in transforming exponentials into simpler linear forms.

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

In 1840 , Britain experienced a bovine (cattle and oxen) epidemic called epizooty. The estimated number of new cases every 28 days is listed in the table. At the time, the London Daily made a dire prediction that the number of new cases would continue to increase indefinitely. William Farr correctly predicted when the number of new cases would peak. Of the two functions $$\begin{array}{l}f(t)=653(1.028)^{t} \\\g(t)=54,700 e^{-(t-200)^{2} / 7500}\end{array}$$ and one models the newspaper's prediction and the other models Farr's prediction, where \(t\) is in days with \(t=0\) corresponding to August \(12,1840\). $$\begin{array}{|c|c|}\hline \text { Date } & \text { New cases } \\\\\hline \text { Aug. 12 } & 506 \\\\\hline \text { Sept. 9 } & 1289 \\\\\hline \text { Oct. 7 } & 3487 \\\\\hline \text { Nov. 4 } & 9597 \\\\\hline \text { Dec. 2 } & 18,817 \\\\\hline \text { Dec. 30 } & 33,835 \\\\\hline \text { Jan. 27 } & 47,191 \\\\\hline\end{array}$$ (a) Graph each function, together with the data, in the viewing rectangle \([0,400,100]\) by \([0,60,000,10,000]\) (b) Determine which function better models Farr's prediction. (c) Determine the date on which the number of new cases peaked.

Exer. \(87-88:\) Graph \(f\) and \(g\) on the same coordinate plane, and estimate the solution of the inequality \(f(x) \geq g(x)\) $$f(x)=x \ln |x|: \quad g(x)=0.15 e^{x}$$

Exer. \(69-70:\) It is suspected that the following data points lie on the graph of \(y=c \log (k x+10),\) where \(c\) and \(k\) are constants. If the data points have three-decimal-place accuracy, is this suspicion correct? $$(0,0.7),(1,0.782),(2,0.847),(4,0.945)$$

Real estate appreciation If the value of real estate increases at a rate of 4\% per year, after \(t\) years the value \(V\) of a house purchased for \(P\) dollars is \(V=P(1.04)^{t} .\) A graph for the value of a house purchased for 80,000 dollars in 1986 is shown in the figure. Approximate the value of the house, to the nearest 1000 dollars, in the year 2016 . (GRAPH CAN'T COPY)

Graph \(f\) on the given interval. (a) Estimate the largest interval \([a, b]\) with \(a< 0< b\) on which \(f\) is one-to-one. (b) If \(g\) is the function with domain \([a, b]\) such that \(g(x)=f(x)\) for \(a \leq x \leq b,\) estimate the domain and range of \(g^{-1}.\) $$f(x)=2.1 x^{3}-2.98 x^{2}-2.11 x+3 ; \quad[-1,2]$$

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.