- #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
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- Tags
- Exponential
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SUMMARY
The discussion focuses on solving the exponential equation involving logarithms: log_{10}[25^{x+1}+4^{x+1}-19\cdot 10^x]=log_{10}[10^{x+1}]. Participants confirm that the solutions are x=0 and x=-2. The equation can be simplified to 25^{x+1}+4^{x+1}=\frac{29}{10}\cdot 10^{x+1} through algebraic manipulation. The conversation highlights the lack of a straightforward method for solving this equation.
- Understanding of logarithmic properties and equations
- Familiarity with exponential functions
- Basic algebraic manipulation skills
- Knowledge of common logarithms (base 10)
- Study the properties of logarithmic functions in detail
- Learn techniques for solving exponential equations
- Explore algebraic manipulation strategies for complex equations
- Investigate the implications of logarithmic identities in problem-solving
Students, mathematicians, and educators interested in solving exponential equations and understanding logarithmic functions will benefit from this discussion.
- #2
mfb
Mentor
- 37,404
- 14,238
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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