The document discusses solving numerical equations involving logarithmic and exponential functions in base 10 or base e. It provides examples of solving log and exponential equations by isolating the part containing the unknown, then rewriting the equation in the opposite form (log to exponential or exponential to log). The key steps outlined are: 1) isolate the exponential/log part containing the unknown, 2) rewrite the equation by "bringing down" exponents as logarithms or vice versa. Several examples are worked through demonstrating these steps.
* Convert from logarithmic to exponential form.
* Convert from exponential to logarithmic form.
* Evaluate logarithms.
* Use common logarithms.
Use natural logarithms.
* Recognize characteristics of parabolas.
* Understand how the graph of a parabola is related to its quadratic function.
* Determine a quadratic function’s minimum or maximum value.
* Solve problems involving a quadratic function’s minimum or maximum value.
This learner's module discusses or talks about the topic of Quadratic Functions. It also discusses what is Quadratic Functions. It also shows how to transform or rewrite the equation f(x)=ax2 + bx + c to f(x)= a(x-h)2 + k. It will also show the different characteristics of Quadratic Functions.
1. Write an equation in standard form of the parabola that has th.docxKiyokoSlagleis
1.
Write an equation in standard form of the parabola that has the same shape as the graph of f(x) = 2x
2
, but with the given point as the vertex (5, 3).
A. f(x) = (2x - 4) + 4
B. f(x) = 2(2x + 8) + 3
C. f(x) = 2(x - 5)
2
+ 3
D. f(x) = 2(x + 3)
2
+ 3
2 of 20
5.0 Points
Find the coordinates of the vertex for the parabola defined by the given quadratic function.
f(x) = 2(x - 3)
2
+ 1
A. (3, 1)
B. (7, 2)
C. (6, 5)
D. (2, 1)
3 of 20
5.0 Points
Find the vertical asymptotes, if any, and the values of x corresponding to holes, if any, of the graph of the following rational function.
g(x) = x + 3/x(x + 4)
A. Vertical asymptotes: x = 4, x = 0; holes at 3x
B. Vertical asymptotes: x = -8, x = 0; holes at x + 4
C. Vertical asymptotes: x = -4, x = 0; no holes
D. Vertical asymptotes: x = 5, x = 0; holes at x - 3
4 of 20
5.0 Points
"Y varies directly as the n
th
power of x" can be modeled by the equation:
A. y = kx
n
.
B. y = kx/n.
C. y = kx
*n
.
D. y = kn
x
.
5 of 20
5.0 Points
40 times a number added to the negative square of that number can be expressed as:
A.
A(x) = x
2
+ 20x.
B. A(x) = -x + 30x.
C.
A(x) = -x
2
- 60x.
D.
A(x) = -x
2
+ 40x.
6 of 20
5.0 Points
The graph of f(x) = -x
3
__________ to the left and __________ to the right.
A. rises; falls
B. falls; falls
C. falls; rises
D. falls; falls
Solve the following formula for the specified variable:
V = 1/3 lwh for h
7 of 20
Write an equation that expresses each relationship. Then solve the equation for y.
x varies jointly as y and z
A. x = kz; y = x/k
B. x = kyz; y = x/kz
C. x = kzy; y = x/z
D. x = ky/z; y = x/zk
8 of 20
8 times a number subtracted from the squared of that number can be expressed as:
A. P(x) = x + 7x.
B.P(x) = x
2
- 8x.
C. P(x) = x - x.
P(x) = x
2
+ 10x.
9of 20
Find the x-intercepts. State whether the graph crosses the x-axis, or touches the x-axis and turns around, at each intercept.
f(x) = x
4
- 9x
2
A. x = 0, x = 3, x = -3; f(x) crosses the x-axis at -3 and 3; f(x) touches the x-axis at 0.
B. x = 1, x = 2, x = 3; f(x) crosses the x-axis at 2 and 3; f(x) crosses the x-axis at 0.
C. x = 0, x = -3, x = 5; f(x) touches the x-axis at -3 and 5; f(x) touches the x-axis at 0.
D. x = 1, x = 2, x = -4; f(x) crosses the x-axis at 2 and -4; f(x) touches the x-axis at 0.
10 of 20
Find the domain of the following rational function.
f(x) = x + 7/x
2
+ 49
A. All real numbers < 69
B. All real numbers > 210
C. All real numbers ≤ 77
D. All real numbers
11 of 20
Write an equation in standard form of the parabola that has the same shape as the graph of f(x) = 3x
2
or g(x) = -3x
2
, but with the given maximum or minimum.
Minimum = 0 at x = 11
A. f(x) = 6(x - 9)
B. f(x) = 3(x - 11)
2
C. f(x) = 4(x + 10)
D. f(x) = 3(x
2
- 15)
2
12 of 20
Solve the following polynomial inequality.
3x
2
+ 10x - 8 ≤ 0
A. [6, 1/3]
B. [-4, 2/3]
C. [-9, 4/5]
D. [8, 2/7]
13 of 20
Find the coordinate.
1) Use properties of logarithms to expand the following logarit.docxhirstcruz
1) Use properties of logarithms to expand the following logarithmic expression as much as possible.
Log
b
(√xy
3
/ z
3
)
A. 1/2 log
b
x - 6 log
b
y + 3 log
b
z
B. 1/2 log
b
x - 9 log
b
y - 3 log
b
z
C. 1/2 log
b
x + 3 log
b
y + 6 log
b
z
D. 1/2 log
b
x + 3 log
b
y - 3 log
b
z
2) Solve the following logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, to two decimal places, for the solution.
2 log x = log 25
A. {12}
B. {5}
C. {-3}
D. {25}
3) Write the following equation in its equivalent logarithmic form.
2
-4
= 1/16
A. Log
4
1/16 = 64
B. Log
2
1/24 = -4
C. Log
2
1/16 = -4
D. Log
4
1/16 = 54
4) Use properties of logarithms to condense the following logarithmic expression. Write the expression as a single logarithm whose coefficient is 1.
log
2
96 – log
2
3
A. 5
B. 7
C. 12
D. 4
5) Use the exponential growth model, A = A
0
e
kt
, to show that the time it takes a population to double (to grow from A
0
to 2A
0
) is given by t = ln 2/k.
A. A
0
= A
0
e
kt
; ln = e
kt
; ln 2 = ln e
kt
; ln 2 = kt; ln 2/k = t
B. 2A
0
= A
0
e; 2= e
kt
; ln = ln e
kt
; ln 2 = kt; ln 2/k = t
C. 2A
0
= A
0
e
kt
; 2= e
kt
; ln 2 = ln e
kt
; ln 2 = kt; ln 2/k = t
D. 2A
0
= A
0
e
kt
; 2 = e
kt
; ln 1 = ln e
kt
; ln 2 = kt; ln 2/k = t
oe
6) Find the domain of following logarithmic function.
f(x) = log (2 - x)
A. (∞, 4)
B. (∞, -12)
C. (-∞, 2)
D. (-∞, -3)
7) An artifact originally had 16 grams of carbon-14 present. The decay model A = 16e -0.000121t describes the amount of carbon-14 present after t years. How many grams of carbon-14 will be present in 5715 years?
A. Approximately 7 grams
B. Approximately 8 grams
C. Approximately 23 grams
D. Approximately 4 grams
8) Use properties of logarithms to expand the following logarithmic expression as much as possible.
log
b
(x
2
y) / z
2
A. 2 log
b
x + log
b
y - 2 log
b
z
B. 4 log
b
x - log
b
y - 2 log
b
z
C. 2 log
b
x + 2 log
b
y + 2 log
b
z
D. log
b
x - log
b
y + 2 log
b
z
9) The exponential function f with base b is defined by f(x) = __________, b > 0 and b ≠ 1. Using interval notation, the domain of this function is __________ and the range is __________.
A. bx; (∞, -∞); (1, ∞)
B. bx; (-∞, -∞); (2, ∞)
C. bx; (-∞, ∞); (0, ∞)
D. bx; (-∞, -∞); (-1, ∞)
10) Approximate the following using a calculator; round your answer to three decimal places.
3
√5
A. .765
B. 14297
C. 11.494
D. 11.665
11) Write the following equation in its equivalent exponential form.
4 = log
2
16
A. 2 log
4
= 16
B. 2
2
= 4
C. 4
4
= 256
D. 2
4
= 16
12) Solve the following exponential equation by expressing each side as a power of the same base and then equating exponents.
3
1-x
= 1/27
A. {2}
B. {-7}
C. {4}
D. {3}
13) Use properties of logarithms to expand the followin.
Epistemic Interaction - tuning interfaces to provide information for AI supportAlan Dix
Paper presented at SYNERGY workshop at AVI 2024, Genoa, Italy. 3rd June 2024
https://alandix.com/academic/papers/synergy2024-epistemic/
As machine learning integrates deeper into human-computer interactions, the concept of epistemic interaction emerges, aiming to refine these interactions to enhance system adaptability. This approach encourages minor, intentional adjustments in user behaviour to enrich the data available for system learning. This paper introduces epistemic interaction within the context of human-system communication, illustrating how deliberate interaction design can improve system understanding and adaptation. Through concrete examples, we demonstrate the potential of epistemic interaction to significantly advance human-computer interaction by leveraging intuitive human communication strategies to inform system design and functionality, offering a novel pathway for enriching user-system engagements.
A tale of scale & speed: How the US Navy is enabling software delivery from l...sonjaschweigert1
Rapid and secure feature delivery is a goal across every application team and every branch of the DoD. The Navy’s DevSecOps platform, Party Barge, has achieved:
- Reduction in onboarding time from 5 weeks to 1 day
- Improved developer experience and productivity through actionable findings and reduction of false positives
- Maintenance of superior security standards and inherent policy enforcement with Authorization to Operate (ATO)
Development teams can ship efficiently and ensure applications are cyber ready for Navy Authorizing Officials (AOs). In this webinar, Sigma Defense and Anchore will give attendees a look behind the scenes and demo secure pipeline automation and security artifacts that speed up application ATO and time to production.
We will cover:
- How to remove silos in DevSecOps
- How to build efficient development pipeline roles and component templates
- How to deliver security artifacts that matter for ATO’s (SBOMs, vulnerability reports, and policy evidence)
- How to streamline operations with automated policy checks on container images
In his public lecture, Christian Timmerer provides insights into the fascinating history of video streaming, starting from its humble beginnings before YouTube to the groundbreaking technologies that now dominate platforms like Netflix and ORF ON. Timmerer also presents provocative contributions of his own that have significantly influenced the industry. He concludes by looking at future challenges and invites the audience to join in a discussion.
Smart TV Buyer Insights Survey 2024 by 91mobiles.pdf91mobiles
91mobiles recently conducted a Smart TV Buyer Insights Survey in which we asked over 3,000 respondents about the TV they own, aspects they look at on a new TV, and their TV buying preferences.
UiPath Test Automation using UiPath Test Suite series, part 4DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 4. In this session, we will cover Test Manager overview along with SAP heatmap.
The UiPath Test Manager overview with SAP heatmap webinar offers a concise yet comprehensive exploration of the role of a Test Manager within SAP environments, coupled with the utilization of heatmaps for effective testing strategies.
Participants will gain insights into the responsibilities, challenges, and best practices associated with test management in SAP projects. Additionally, the webinar delves into the significance of heatmaps as a visual aid for identifying testing priorities, areas of risk, and resource allocation within SAP landscapes. Through this session, attendees can expect to enhance their understanding of test management principles while learning practical approaches to optimize testing processes in SAP environments using heatmap visualization techniques
What will you get from this session?
1. Insights into SAP testing best practices
2. Heatmap utilization for testing
3. Optimization of testing processes
4. Demo
Topics covered:
Execution from the test manager
Orchestrator execution result
Defect reporting
SAP heatmap example with demo
Speaker:
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
The Art of the Pitch: WordPress Relationships and SalesLaura Byrne
Clients don’t know what they don’t know. What web solutions are right for them? How does WordPress come into the picture? How do you make sure you understand scope and timeline? What do you do if sometime changes?
All these questions and more will be explored as we talk about matching clients’ needs with what your agency offers without pulling teeth or pulling your hair out. Practical tips, and strategies for successful relationship building that leads to closing the deal.
Securing your Kubernetes cluster_ a step-by-step guide to success !KatiaHIMEUR1
Today, after several years of existence, an extremely active community and an ultra-dynamic ecosystem, Kubernetes has established itself as the de facto standard in container orchestration. Thanks to a wide range of managed services, it has never been so easy to set up a ready-to-use Kubernetes cluster.
However, this ease of use means that the subject of security in Kubernetes is often left for later, or even neglected. This exposes companies to significant risks.
In this talk, I'll show you step-by-step how to secure your Kubernetes cluster for greater peace of mind and reliability.
Threats to mobile devices are more prevalent and increasing in scope and complexity. Users of mobile devices desire to take full advantage of the features
available on those devices, but many of the features provide convenience and capability but sacrifice security. This best practices guide outlines steps the users can take to better protect personal devices and information.
State of ICS and IoT Cyber Threat Landscape Report 2024 previewPrayukth K V
The IoT and OT threat landscape report has been prepared by the Threat Research Team at Sectrio using data from Sectrio, cyber threat intelligence farming facilities spread across over 85 cities around the world. In addition, Sectrio also runs AI-based advanced threat and payload engagement facilities that serve as sinks to attract and engage sophisticated threat actors, and newer malware including new variants and latent threats that are at an earlier stage of development.
The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
Download the full report from here:
https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
DevOps and Testing slides at DASA ConnectKari Kakkonen
My and Rik Marselis slides at 30.5.2024 DASA Connect conference. We discuss about what is testing, then what is agile testing and finally what is Testing in DevOps. Finally we had lovely workshop with the participants trying to find out different ways to think about quality and testing in different parts of the DevOps infinity loop.
Transcript: Selling digital books in 2024: Insights from industry leaders - T...BookNet Canada
The publishing industry has been selling digital audiobooks and ebooks for over a decade and has found its groove. What’s changed? What has stayed the same? Where do we go from here? Join a group of leading sales peers from across the industry for a conversation about the lessons learned since the popularization of digital books, best practices, digital book supply chain management, and more.
Link to video recording: https://bnctechforum.ca/sessions/selling-digital-books-in-2024-insights-from-industry-leaders/
Presented by BookNet Canada on May 28, 2024, with support from the Department of Canadian Heritage.
Removing Uninteresting Bytes in Software FuzzingAftab Hussain
Imagine a world where software fuzzing, the process of mutating bytes in test seeds to uncover hidden and erroneous program behaviors, becomes faster and more effective. A lot depends on the initial seeds, which can significantly dictate the trajectory of a fuzzing campaign, particularly in terms of how long it takes to uncover interesting behaviour in your code. We introduce DIAR, a technique designed to speedup fuzzing campaigns by pinpointing and eliminating those uninteresting bytes in the seeds. Picture this: instead of wasting valuable resources on meaningless mutations in large, bloated seeds, DIAR removes the unnecessary bytes, streamlining the entire process.
In this work, we equipped AFL, a popular fuzzer, with DIAR and examined two critical Linux libraries -- Libxml's xmllint, a tool for parsing xml documents, and Binutil's readelf, an essential debugging and security analysis command-line tool used to display detailed information about ELF (Executable and Linkable Format). Our preliminary results show that AFL+DIAR does not only discover new paths more quickly but also achieves higher coverage overall. This work thus showcases how starting with lean and optimized seeds can lead to faster, more comprehensive fuzzing campaigns -- and DIAR helps you find such seeds.
- These are slides of the talk given at IEEE International Conference on Software Testing Verification and Validation Workshop, ICSTW 2022.
Essentials of Automations: The Art of Triggers and Actions in FMESafe Software
In this second installment of our Essentials of Automations webinar series, we’ll explore the landscape of triggers and actions, guiding you through the nuances of authoring and adapting workspaces for seamless automations. Gain an understanding of the full spectrum of triggers and actions available in FME, empowering you to enhance your workspaces for efficient automation.
We’ll kick things off by showcasing the most commonly used event-based triggers, introducing you to various automation workflows like manual triggers, schedules, directory watchers, and more. Plus, see how these elements play out in real scenarios.
Whether you’re tweaking your current setup or building from the ground up, this session will arm you with the tools and insights needed to transform your FME usage into a powerhouse of productivity. Join us to discover effective strategies that simplify complex processes, enhancing your productivity and transforming your data management practices with FME. Let’s turn complexity into clarity and make your workspaces work wonders!
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
SAP Sapphire 2024 - ASUG301 building better apps with SAP Fiori.pdfPeter Spielvogel
Building better applications for business users with SAP Fiori.
• What is SAP Fiori and why it matters to you
• How a better user experience drives measurable business benefits
• How to get started with SAP Fiori today
• How SAP Fiori elements accelerates application development
• How SAP Build Code includes SAP Fiori tools and other generative artificial intelligence capabilities
• How SAP Fiori paves the way for using AI in SAP apps
2. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e.
Calculation with Log and Exp
3. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e. Most numerical calculations in science are
in these two bases.
Calculation with Log and Exp
4. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e. Most numerical calculations in science are
in these two bases. We need a calculator that has
the following functions: ex, 10x, ln(x), and log(x).
Calculation with Log and Exp
5. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e. Most numerical calculations in science are
in these two bases. We need a calculator that has
the following functions: ex, 10x, ln(x), and log(x).
All answers are given to 3 significant digits.
Calculation with Log and Exp
6. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e. Most numerical calculations in science are
in these two bases. We need a calculator that has
the following functions: ex, 10x, ln(x), and log(x).
All answers are given to 3 significant digits.
6
Example A: Find the answers with a calculator.
a.103.32 b. e = e1/6
c. log(4.35) d. ln(2/3)
Calculation with Log and Exp
7. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e. Most numerical calculations in science are
in these two bases. We need a calculator that has
the following functions: ex, 10x, ln(x), and log(x).
All answers are given to 3 significant digits.
6
Example A: Find the answers with a calculator.
a.103.32 b. e = e1/6
2090
c. log(4.35) d. ln(2/3)
Calculation with Log and Exp
8. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e. Most numerical calculations in science are
in these two bases. We need a calculator that has
the following functions: ex, 10x, ln(x), and log(x).
All answers are given to 3 significant digits.
6
Example A: Find the answers with a calculator.
a.103.32 b. e = e1/6
2090 1.18
c. log(4.35) d. ln(2/3)
Calculation with Log and Exp
9. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e. Most numerical calculations in science are
in these two bases. We need a calculator that has
the following functions: ex, 10x, ln(x), and log(x).
All answers are given to 3 significant digits.
6
Example A: Find the answers with a calculator.
a.103.32 b. e = e1/6
2090 1.18
c. log(4.35) d. ln(2/3)
0.638
Calculation with Log and Exp
10. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e. Most numerical calculations in science are
in these two bases. We need a calculator that has
the following functions: ex, 10x, ln(x), and log(x).
All answers are given to 3 significant digits.
6
Example A: Find the answers with a calculator.
a.103.32 b. e = e1/6
2090 1.18
c. log(4.35) d. ln(2/3)
0.638 -0.405
Calculation with Log and Exp
11. In this section, we solve simple numerical equations
involving log and exponential functions in base 10
or base e. Most numerical calculations in science are
in these two bases. We need a calculator that has
the following functions: ex, 10x, ln(x), and log(x).
All answers are given to 3 significant digits.
6
Example A: Find the answers with a calculator.
a.103.32 b. e = e1/6
2090 1.18
c. log(4.35) d. ln(2/3)
0.638 -0.405
These problems may be stated in alternate forms.
Calculation with Log and Exp
12. Example B: Find the x
a. log(x) = 3.32 b. 1/6 = ln(x)
c. 10x = 4.35 d. 2/3 = ex
Calculation with Log and Exp
13. Example B: Find the x
a. log(x) = 3.32 b. 1/6 = ln(x)
x =103.32 ( 2090)
c. 10x = 4.35 d. 2/3 = ex
Calculation with Log and Exp
14. Example B: Find the x
a. log(x) = 3.32 b. 1/6 = ln(x)
x =103.32 ( 2090) e1/6 = x ( 1.18)
c. 10x = 4.35 d. 2/3 = ex
Calculation with Log and Exp
15. Example B: Find the x
a. log(x) = 3.32 b. 1/6 = ln(x)
x =103.32 ( 2090) e1/6 = x ( 1.18)
c. 10x = 4.35 d. 2/3 = ex
x = log(4.35) ( 0.638)
Calculation with Log and Exp
16. Example B: Find the x
a. log(x) = 3.32 b. 1/6 = ln(x)
x =103.32 ( 2090) e1/6 = x ( 1.18)
c. 10x = 4.35 d. 2/3 = ex
x = log(4.35) ( 0.638) ln(2/3) = x ( -0.405)
Calculation with Log and Exp
17. Example B: Find the x
a. log(x) = 3.32 b. 1/6 = ln(x)
x =103.32 ( 2090) e1/6 = x ( 1.18)
c. 10x = 4.35 d. 2/3 = ex
x = log(4.35) ( 0.638) ln(2/3) = x ( -0.405)
An equation is called a log-equations if the unknown
is in the log-function as in parts a and b above.
Calculation with Log and Exp
18. Example B: Find the x
a. log(x) = 3.32 b. 1/6 = ln(x)
x =103.32 ( 2090) e1/6 = x ( 1.18)
c. 10x = 4.35 d. 2/3 = ex
x = log(4.35) ( 0.638) ln(2/3) = x ( -0.405)
An equation is called an exponential equations if the
unknown is in the exponent as in parts c and d.
An equation is called a log-equations if the unknown
is in the log-function as in parts a and b above.
Calculation with Log and Exp
19. Example B: Find the x
a. log(x) = 3.32 b. 1/6 = ln(x)
x =103.32 ( 2090) e1/6 = x ( 1.18)
c. 10x = 4.35 d. 2/3 = ex
x = log(4.35) ( 0.638) ln(2/3) = x ( -0.405)
An equation is called an exponential equations if the
unknown is in the exponent as in parts c and d.
An equation is called a log-equations if the unknown
is in the log-function as in parts a and b above.
To solve log-equations, drop the log and write the
problems in exp-form.
Calculation with Log and Exp
20. Example B: Find the x
a. log(x) = 3.32 b. 1/6 = ln(x)
x =103.32 ( 2090) e1/6 = x ( 1.18)
c. 10x = 4.35 d. 2/3 = ex
x = log(4.35) ( 0.638) ln(2/3) = x ( -0.405)
An equation is called an exponential equations if the
unknown is in the exponent as in parts c and d.
An equation is called a log-equations if the unknown
is in the log-function as in parts a and b above.
To solve log-equations, drop the log and write the
problems in exp-form. To solve exponential
equations, lower the exponents and write the
problems in log-form.
Calculation with Log and Exp
21. More precisely, to solve exponential equations,
Calculation with Log and Exp
22. More precisely, to solve exponential equations, we
I. isolate the exponential part that contains the x,
Calculation with Log and Exp
23. More precisely, to solve exponential equations, we
I. isolate the exponential part that contains the x,
II. bring down the exponents by writing it in log-form.
Calculation with Log and Exp
24. More precisely, to solve exponential equations, we
I. isolate the exponential part that contains the x,
II. bring down the exponents by writing it in log-form.
Example C: Solve 25 = 7*102x
Calculation with Log and Exp
25. More precisely, to solve exponential equations, we
I. isolate the exponential part that contains the x,
II. bring down the exponents by writing it in log-form.
Example C: Solve 25 = 7*102x
Isolate the exponential part containing the x,
25/7 = 102x
Calculation with Log and Exp
26. More precisely, to solve exponential equations, we
I. isolate the exponential part that contains the x,
II. bring down the exponents by writing it in log-form.
Example C: Solve 25 = 7*102x
Isolate the exponential part containing the x,
25/7 = 102x
Bring down the x by restating it in log-form:
log(25/7) = 2x
Calculation with Log and Exp
27. More precisely, to solve exponential equations, we
I. isolate the exponential part that contains the x,
II. bring down the exponents by writing it in log-form.
Example C: Solve 25 = 7*102x
Isolate the exponential part containing the x,
25/7 = 102x
Bring down the x by restating it in log-form:
log(25/7) = 2x
log(25/7)
2
= x
Exact answer
Calculation with Log and Exp
28. More precisely, to solve exponential equations, we
I. isolate the exponential part that contains the x,
II. bring down the exponents by writing it in log-form.
Example C: Solve 25 = 7*102x
Isolate the exponential part containing the x,
25/7 = 102x
Bring down the x by restating it in log-form:
log(25/7) = 2x
log(25/7)
2
= x 0.276
Exact answer Approx. answer
Calculation with Log and Exp
29. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Calculation with Log and Exp
30. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Isolate the exp-part: 2.3*e2-3x + 4.1 = 12.5
Calculation with Log and Exp
31. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Isolate the exp-part: 2.3*e2-3x + 4.1 = 12.5
2.3*e2-3x = 12.5 – 4.1
2.3*e2-3x = 8.4
Calculation with Log and Exp
32. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Isolate the exp-part: 2.3*e2-3x + 4.1 = 12.5
2.3*e2-3x = 12.5 – 4.1
2.3*e2-3x = 8.4
e2-3x = 8.4/2.3
Calculation with Log and Exp
33. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Isolate the exp-part: 2.3*e2-3x + 4.1 = 12.5
2.3*e2-3x = 12.5 – 4.1
2.3*e2-3x = 8.4
e2-3x = 8.4/2.3
Restate in log-form: 2 – 3x = ln(8.4/2.3)
Calculation with Log and Exp
34. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Isolate the exp-part: 2.3*e2-3x + 4.1 = 12.5
2.3*e2-3x = 12.5 – 4.1
2.3*e2-3x = 8.4
e2-3x = 8.4/2.3
Restate in log-form: 2 – 3x = ln(8.4/2.3)
Solve for x: 2 – ln(8.4/2.3) = 3x
Calculation with Log and Exp
35. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Isolate the exp-part: 2.3*e2-3x + 4.1 = 12.5
2.3*e2-3x = 12.5 – 4.1
2.3*e2-3x = 8.4
e2-3x = 8.4/2.3
Restate in log-form: 2 – 3x = ln(8.4/2.3)
Solve for x: 2 – ln(8.4/2.3) = 3x
2-ln(8.4/2.3)
3
= x
Calculation with Log and Exp
36. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Isolate the exp-part: 2.3*e2-3x + 4.1 = 12.5
2.3*e2-3x = 12.5 – 4.1
2.3*e2-3x = 8.4
e2-3x = 8.4/2.3
Restate in log-form: 2 – 3x = ln(8.4/2.3)
Solve for x: 2 – ln(8.4/2.3) = 3x
2-ln(8.4/2.3)
3
= x 0.235
Calculation with Log and Exp
37. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Isolate the exp-part: 2.3*e2-3x + 4.1 = 12.5
2.3*e2-3x = 12.5 – 4.1
2.3*e2-3x = 8.4
e2-3x = 8.4/2.3
Restate in log-form: 2 – 3x = ln(8.4/2.3)
Solve for x: 2 – ln(8.4/2.3) = 3x
2-ln(8.4/2.3)
3
= x 0.235
Calculation with Log and Exp
We solve log-equation in analogous fashion:
38. Example D: Solve 2.3*e2-3x + 4.1 = 12.5
Isolate the exp-part: 2.3*e2-3x + 4.1 = 12.5
2.3*e2-3x = 12.5 – 4.1
2.3*e2-3x = 8.4
e2-3x = 8.4/2.3
Restate in log-form: 2 – 3x = ln(8.4/2.3)
Solve for x: 2 – ln(8.4/2.3) = 3x
2-ln(8.4/2.3)
3
= x 0.235
Calculation with Log and Exp
We solve log-equation in analogous fashion:
I. isolate the log part that contains the x,
II. drop the log by writing it in exp-form.
40. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
41. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
42. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x:
43. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x: 2x = 107/9 – 1
x = (107/9 – 1)/2
44. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x: 2x = 107/9 – 1
x = (107/9 – 1)/2 2.50
45. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x: 2x = 107/9 – 1
x = (107/9 – 1)/2 2.50
Example F: Solve 2.3*log(2–3x)+4.1 = 12.5
46. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x: 2x = 107/9 – 1
x = (107/9 – 1)/2 2.50
Example F: Solve 2.3*log(2–3x)+4.1 = 12.5
2.3*log(2–3x) + 4.1 = 12.5
47. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x: 2x = 107/9 – 1
x = (107/9 – 1)/2 2.50
Example F: Solve 2.3*log(2–3x)+4.1 = 12.5
2.3*log(2–3x) + 4.1 = 12.5
2.3*log(2–3x) = 12.5 – 4.1
2.3*log(2–3x) = 8.4
48. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x: 2x = 107/9 – 1
x = (107/9 – 1)/2 2.50
Example F: Solve 2.3*log(2–3x)+4.1 = 12.5
2.3*log(2–3x) + 4.1 = 12.5
2.3*log(2–3x) = 12.5 – 4.1
2.3*log(2–3x) = 8.4
log(2 – 3x) = 8.4/2.3
49. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x: 2x = 107/9 – 1
x = (107/9 – 1)/2 2.50
Example F: Solve 2.3*log(2–3x)+4.1 = 12.5
2.3*log(2–3x) + 4.1 = 12.5
2.3*log(2–3x) = 12.5 – 4.1
2.3*log(2–3x) = 8.4
log(2 – 3x) = 8.4/2.3
2 – 3x = 108.4/2.3
50. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x: 2x = 107/9 – 1
x = (107/9 – 1)/2 2.50
Example F: Solve 2.3*log(2–3x)+4.1 = 12.5
2.3*log(2–3x) + 4.1 = 12.5
2.3*log(2–3x) = 12.5 – 4.1
2.3*log(2–3x) = 8.4
log(2 – 3x) = 8.4/2.3
2 – 3x = 108.4/2.3
2 – 108.4/2.3 = 3x
51. Calculation with Log and Exp
Example E: Solve 9*log(2x+1)= 7
Isolate the log-part, log(2x+1) = 7/9
Write it in exp-form 2x + 1 = 107/9
Save for x: 2x = 107/9 – 1
x = (107/9 – 1)/2 2.50
Example F: Solve 2.3*log(2–3x)+4.1 = 12.5
2.3*log(2–3x) + 4.1 = 12.5
2.3*log(2–3x) = 12.5 – 4.1
2.3*log(2–3x) = 8.4
log(2 – 3x) = 8.4/2.3
2 – 3x = 108.4/2.3
2 – 108.4/2.3 = 3x
2 – 108.4/2.3
= x -14953
52. Solve the following exponential equations, give the exact and the approximate solutions.
1. 5e2x = 7 2. 3e - 2x+1 = 6
Exact answer: x = ½* LN(7/5) Exact answer: x = (1 – LN(2)) /2
Aproxímate: 0.168 Aproxímate: 0.153
3. 4 – e 3x+ 1 = 2 4. 2* 10 3x - 2 = 5
Exact answer: x = (LN(2) – 1)/3 Exact answer: x = (LOG(5/2) + 2)/3
Approximate: - 0.102 Approximate: 0.799
5. 6 + 3* 10 1- x = 10 6. -7 – 3*10 2x - 1 = -24
Exact answer: x = 1 – LOG(4/3) Exact answer: x = (LOG(17/3)+1)/2
Aproxímate: 0.875 Aproxímate: 0.877
7. 8 = 12 – 2e 2- x 8. 5*10 2 - 3x + 3 = 14
Exact answer: x = 2 – LN(2) Exact answer: x = (2 – LOG(11/5)) /3
Approximate: 1.31 Approximate: 0.553
Solve the following log equations, give the exact and the approximate solutions.
9. LOG(3x+1) = 3/5 10. LN(2 – x) = -2/3
Exact answer: x = (103/5 – 1)/3 Exact answer: x = 2 – e -2/3
Approximate: 0.994 Approximate: 1.49
11. 2LOG(2x –3) = 1/3 12. 2 + Log(4 – 2x) = -8
Exact answer: x = (101/6 + 3)/2 Exact answer: x = (4 – 10-10)/2
Approximate: 2.23 Approximate: 2.000
13. 3 – 5LN(3x +1) = -8 14. -3 +5LOG(1 – 2x) = 9
Exact answer: x = (e11/5 – 1 )/3 Exact answer: x = (1 – 10 12/5)/2
Approximate: 2.68 Approximate: -125
15. 2LN(2x – 1) – 3 = 5 16. 7 – 2LN(12x+15) =23
Exact answer: x = (e4+1)/2 Exact answer: x = (e-8 – 15 )/12
Approximate: 27.8 Approximate: -1.25