How to Solve Exponential Equations with Natural Logs?

  • Thread starter Thread starter pavadrin
  • Start date Start date
  • Tags Tags
    Natural Test
Click For Summary

Homework Help Overview

The problem involves solving an exponential equation, specifically 50e^(2x) = 100e^(0.5x), within the context of introductory calculus. The original poster expresses frustration with the problem, particularly regarding the application of natural logarithms and the coefficients involved.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the potential of manipulating the equation by multiplying both sides by exp(-0.5x) and explore the implications of this step. Some share their attempts at solving the equation and express confusion about the process.

Discussion Status

The discussion includes various approaches to the problem, with some participants questioning the effectiveness of their methods. While there are attempts to guide understanding through manipulation of the equation, there is no clear consensus on the best approach yet.

Contextual Notes

Participants note the challenge posed by the coefficients in the equation and the requirement to use natural logarithms, indicating a level of complexity in the problem that may affect their understanding.

pavadrin
Messages
154
Reaction score
0
hey
2day i had a test in intro calc and was given the following problem to solve:

[tex]50e^(2x) = 100e^(0.5x)[/tex]

i was quite annoyed :mad: at this problem because i didn’t know how to solve it. although i do know that natural logs need to be taken on both sides to solve it, however the coefficient of the e really had me confused. any help would be greatly appreciated, thanks,
Pavadrin
 
Physics news on Phys.org
What happens if you multiply both sides by exp(-0.5x)... ?
 
I got stuck on the same question :(

heres the working i did for it after the test.

http://img145.imageshack.us/img145/3503/logszp0.gif
 
Last edited by a moderator:
if both sides are multiplied by exp(-0.5x) would that make it:

[tex]50e^2.5x = 100e^x[/tex]

but i don't see how that helps, thanks for the relpy though
 
eax * e-ax = eax-ax = e0 = 1

danago is correct.

Using ln,

[tex]50e^{2x} = 100e^{0.5x}[/tex]

start by dividing the equation by 50

[tex]e^{2x} = 2e^{0.5x}[/tex]

take natural log

[tex]2x = ln 2\,+\,0.5x[/tex]

[tex]1.5x = ln 2[/tex]

[tex]x = \frac{ln 2}{1.5}[/tex]
 
Last edited:
Astronuc said:
danago is correct.
I just didn't want to give the game away :smile:
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
18
Views
4K
Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 26 ·
Replies
26
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K