Can someone double check my work in this problem?

  • Thread starter Thread starter name_ask17
  • Start date Start date
  • Tags Tags
    Check my work Work
AI Thread Summary
The user seeks confirmation on their factoring of two expressions: 2e^(3x) - 5xe^(3x) and 4ln(x) - 3yln(x). They factored the first as e^(3x)(2 - 5x) and the second as ln(x)(4 - 3y). Responses confirm both factorizations are correct but suggest using parentheses for clarity, particularly in the expression involving the natural logarithm. The discussion emphasizes the importance of notation to avoid misunderstandings. Overall, the user's work is validated with minor recommendations for improved clarity.
name_ask17
Messages
146
Reaction score
0

Homework Statement



Okay, I think these are simple, but just so I know that I am on the right track, can you double check my work? My questions and answers are in bold.

Factor these completely:

1. 2e^3x - 5xe^3x

2. 4lnx -3ylnx


3. The Attempt at a Solution

1. I took out the e^3x to make the factored problem e^3x(2-5x)Did I do that right, because I get confused when dealing with the exponents. If there is more to it, can you explain please? Thanks in advance.

2. I just took out the lnx to make it lnx(4-3y) . Is this correct?
 
Physics news on Phys.org
name_ask17 said:

Homework Statement



Okay, I think these are simple, but just so I know that I am on the right track, can you double check my work? My questions and answers are in bold.

Factor these completely:

1. 2e^3x - 5xe^3x

2. 4lnx -3ylnx


3. The Attempt at a Solution

1. I took out the e^3x to make the factored problem e^3x(2-5x)Did I do that right, because I get confused when dealing with the exponents. If there is more to it, can you explain please? Thanks in advance.

2. I just took out the lnx to make it lnx(4-3y) . Is this correct?

They are both correct but you need parentheses. For example e^(3x). Also, while not incorrect, it is better to write the last one (4-3y)ln(x) to remove any possible misunderstanding about what is inside the ln function.
 
Thanks! Just wanted to make sure they were correct because it seemed too simple. Guess I'm used to overthinking things. lol.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top