# 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

#### 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? Last edited:

#### 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.

### Want to reply to this thread?

"Exponential equation with logs"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving