How to Solve a Complex Algebraic Equation with Logs and Square Roots?

  • Context: High School 
  • Thread starter Thread starter Matrix
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around solving a complex algebraic equation involving logarithms and square roots. Participants explore different methods to find the variable x, including the use of logarithms and alternative approaches.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant presents the equation 2x = 16 - 3/4 * sqrt(2) and seeks assistance in solving for x.
  • Another participant suggests using logarithms to solve the equation, transforming it into the form 2^x = 16^(-3/4) * sqrt(2) and deriving x = ln(16^(-3/4) * sqrt(2)) / ln(2).
  • A subsequent post claims that the solution simplifies to x = -2.5.
  • One participant challenges the correctness of the reduction made by another, indicating a potential error in the previous steps.
  • Another participant questions whether it is possible to solve for x without using logarithms.
  • A participant asserts that it is indeed possible to solve without logs by converting 16 to a base of 2, leading to the conclusion that x = -2.5.
  • A later post expresses a fondness for logarithms, suggesting a personal preference rather than a technical point.

Areas of Agreement / Disagreement

There is disagreement regarding the correctness of the initial reduction steps, with some participants asserting different views on the validity of using logarithms versus alternative methods. The discussion remains unresolved on the best approach to solving the equation.

Contextual Notes

Participants express uncertainty about the correctness of certain mathematical reductions and the applicability of different methods, indicating a reliance on specific transformations and definitions.

Matrix
Messages
4
Reaction score
0
Hi, I am having trouble finding x for the following exquation:
2x=16-3/4 * sqrt(2)

Any help on solving it would be much apprecitated.
 
Last edited:
Mathematics news on Phys.org
why don't you use logs.

2^x = 16^(-3/4) * sqrt(2)

x*ln(2) = ln(16^(-3/4) * sqrt(2) )

x = ln(16^(-3/4) * sqrt(2) ) / ln(2)
 
reduce it further to:

x = -2.5
 
jcsd, your i see you changed your reduction. The previous one was obviously wrong.
 
Is it possible to find x not using logs?
 
Duardo: No, it was perfectly correct.
 
Originally posted by Matrix
Is it possible to find x not using logs?

yes.

you can convert 16 to a base 2

16^(-3/4) = (2^4)^(-3/4) = 2^(-3)

therefore

2^x = 2^(-3) * 2^(1/2)

2^x = 2^(-5/2)

x = -2.5
 
This is probably off topic, but don't you just love logs?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 41 ·
2
Replies
41
Views
7K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 44 ·
2
Replies
44
Views
5K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K