# I Exponential equation with logs

#### Dank2

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

#### mfb

Mentor
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.

#### Dank2

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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#### 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.

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