- #1
Dank2
- 213
- 4
How can i continue from here, answer is x=0,-2
![IMG_4519[894].JPG](https://www.physicsforums.com/attachments/img_4519-894-jpg.249209/ "IMG_4519[894].JPG")
Exponential equation with logs
- Context: Undergrad
- Thread starter Dank2
- Start date
-
- Tags
- Exponential
Click For Summary
Discussion Overview
The discussion revolves around solving an exponential equation involving logarithms. Participants explore various approaches to find the solutions, which include specific values for x, and engage in algebraic manipulation of the equation.
Discussion Character
- Mathematical reasoning, Debate/contested
Main Points Raised
- One participant suggests that the solutions to the equation are x=0 and x=-2, but seeks further clarification on how to demonstrate this.
- Another participant points out a potential error regarding a missing factor in the equation, suggesting that common factors of 10^x could be useful for simplification.
- There is a repeated mention of the solutions being x=0 and x=-1, with a request for a method to show this without needing a formal proof.
- A participant provides a specific form of the equation and indicates that algebraic manipulation can lead to a clearer understanding of the solutions.
- One participant dismisses a value of t=0 as irrelevant to the discussion.
Areas of Agreement / Disagreement
Participants express differing views on the solutions, with some proposing x=0 and x=-2, while others suggest x=0 and x=-1. The discussion remains unresolved regarding the correct solutions and the methods to demonstrate them.
Contextual Notes
There are indications of missing assumptions and potential errors in the manipulation of the equation, as well as unresolved steps in the algebraic process. The scope of the discussion is limited to finding solutions without formal proofs.
- #2
mfb
Mentor
- 37,418
- 14,253
You lost a factor 4 for 2x 2x.
I would make some common factors of 10x and from there on say that you can see the solution. There is no nice way to solve this as far as I can see.
I would make some common factors of 10x and from there on say that you can see the solution. There is no nice way to solve this as far as I can see.
- #3
Dank2
- 213
- 4
You mean write down 25 as 10log1025?mfb said:
Last edited:
- #4
Dank2
- 213
- 4
dont mind the t=0, since it cannot be.
- #5
Dank2
- 213
- 4
Answers are x=0, x=-1.
Anyone have a clue how to show it ? it doesn't have to be a proof.
Anyone have a clue how to show it ? it doesn't have to be a proof.
- #6
aheight
- 318
- 108
If the equation is:Dank2 said:
$$\log_{10}\big[25^{x+1}+4^{x+1}-19\cdot 10^x\big]=\log_{10}\big[10^{x+1}\big]$$,
then with some algebra, you can reduce it down to:
$$
25^{x+1}+4^{x+1}=\frac{29}{10}\cdot 10^{x+1}
$$
and can easily see ##x=0## is a solution and if you plug in ##x=-2##, confirm -2 is a solution also.
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