Exponential equation with logs

[Dank2]

How can i continue from here, answer is x=0,-2

 

[mfb]

You lost a factor 4 for 2^x 2^x.

I would make some common factors of 10^x 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.

 

[Dank2]

You lost a factor 4 for 2^x 2^x.

I would make some common factors of 10^x 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 10^log₁₀25?

 

[Dank2]

dont mind the t=0, since it cannot be.

 

[Dank2]

Answers are x=0, x=-1.
Anyone have a clue how to show it ? it doesn't have to be a proof.

 

[aheight]

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.