I Exponential equation with logs

  • Thread starter Dank2
  • Start date
213
4
How can i continue from here, answer is x=0,-2

IMG_4519[894].JPG
 
33,236
8,950
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.
 
213
4
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?
D356CA16-1260-4F12-83CB-98BEBAAF25EC.jpeg
 
Last edited:
213
4
dont mind the t=0, since it cannot be.
66EFA9ED-15A7-4120-BB0D-5871FF852226.jpeg
 
213
4
Answers are x=0, x=-1.
Anyone have a clue how to show it ? it doesn't have to be a proof.
 
203
46
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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