I Exponential equation with logs

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The discussion revolves around solving the equation involving logarithms and exponential terms, specifically $$\log_{10}\big[25^{x+1}+4^{x+1}-19\cdot 10^x\big]=\log_{10}\big[10^{x+1}\big]$$. Participants note that the equation can be simplified to $$25^{x+1}+4^{x+1}=\frac{29}{10}\cdot 10^{x+1}$$, leading to solutions of x=0 and x=-2. There is a consensus that the problem lacks a straightforward solution method. The conversation highlights the importance of recognizing common factors and algebraic manipulation in reaching the answers. Overall, the key solutions identified are x=0 and x=-2.
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How can i continue from here, answer is x=0,-2

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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.
 
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mfb said:
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.
You mean write down 25 as 10log1025?
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dont mind the t=0, since it cannot be.
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Answers are x=0, x=-1.
Anyone have a clue how to show it ? it doesn't have to be a proof.
 
Dank2 said:
Answers are x=0, x=-1.
Anyone have a clue how to show it ? it doesn't have to be a proof.
If the equation is:
$$\log_{10}\big[25^{x+1}+4^{x+1}-19\cdot 10^x\big]=\log_{10}\big[10^{x+1}\big]$$,
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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