How Do I Solve a Logarithmic Equation with Different Bases and Variables?

  • Context: MHB 
  • Thread starter Thread starter Monoxdifly
  • Start date Start date
  • Tags Tags
    Logarithmic
Click For Summary

Discussion Overview

The discussion revolves around solving a logarithmic equation involving different bases and variables, specifically the equation $$log_2(x+2)+log_{(x-2)}4=3$$. Participants explore the challenges of solving this equation and share their reasoning and calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant expresses uncertainty about solving the equation due to the presence of different variables in the logarithmic terms.
  • The same participant attempts to manipulate the equation using logarithmic identities but questions whether their steps are correct.
  • Another participant suggests that no value of x satisfies the equation, referencing graphical analysis.
  • A subsequent post reiterates the claim that no x fulfills the equation, again based on graphical evidence.
  • A final post acknowledges the contributions of others without further elaboration.

Areas of Agreement / Disagreement

There is no consensus on whether a solution exists, as some participants suggest that no x fulfills the equation while others have not confirmed this conclusion.

Contextual Notes

The discussion includes manipulations of logarithmic expressions that may depend on specific assumptions about the values of x, which are not fully explored or resolved.

Monoxdifly
MHB
Messages
288
Reaction score
0
A friend asked me how to solve this question:
$$log_2(x+2)+log_{(x-2)}4=3$$
I said I had no idea because one is x + 2 and the other one is x - 2. If both are x + 2 or x - 2, I can do it. He said that if that's the case, even at his level he could solve it. This is what I've done so far regarding the question.
$$log_2(x+2)+log_{(x-2)}4=3$$
$$\frac{log(x+2)}{log2}+\frac{log4}{log(x-2)}=3$$
$$\frac{log(x+2)log(x-2)+log4log2}{log2log(x-2)}=3$$
$$log(x+2)log(x-2)+2log^22=3log2log(x-2)$$
$$log(x+2)log(x-2)-3log2log(x-2)=-2log^22$$
$$log(x-2)(log(x+2)-3log2)=-2log^22$$
What should I do from here? Or did I make some mistakes?
 
Mathematics news on Phys.org
Beer hangover induced idea follows.
20200915_141613.jpg
 
So, no x fulfills the equation, right?
 
Monoxdifly said:
So, no x fulfills the equation, right?

That's what the graph says ...
 
Okay, thanks guys.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K