# please try, how is it?

Discussion in 'Basic Math' started by hall, Aug 15, 2023.

1. ### hall

Joined:
Aug 15, 2023
Messages:
1
0
(5^7) (5^x+5^2)=5^16+5^x=(5^x)(5^7)+5^9 then 5^x(1-5^7)=5^9-5^16
then 5^x=(5^16-5^9)/(5^7-5^0)=(5^9-5^2)/(1-1/(5^7))=?
5^a=(5^b-5^c)/(1-1/(5^ (b-c)))=? then a=?b=?9
1 < c < b then each integer b and c then
5 ^ b=?(5 ^ b-5 ^ c)/(1-1 /(5^(b-c)))
c = 2 and b = 3 ok
c = 2 and b = 4 ok
c =3 and b = 4 ok
c = 2 and b = 5 ok
c = 3 and b = 5 ok
c = 4 and b = 5 ok
...
c =5 and b = 6 ok
...
c = 23 and b = 29 ok
...
think deeply and try proof please:
1 < c < b then each integer b and c then
5 ^ b=?(5 ^ b-5 ^ c)/(1-1 /(5^(b-c)))
i can see this, but I can't prove it yet!

hall, Aug 15, 2023
2. ### e.jane.aran

Joined:
Aug 13, 2023
Messages:
32
28
I *think* the following is what is in your post:

===============================

Then:

Then:

= ??

= ??

===============================

I'm not sure what is meant by the rest of your post...? Are you proposing a formula, and are asking for assistance in proving it? If not, what (precisely) is your question?

Thanks!

#### Attached Files:

File size:
1.5 KB
Views:
9
e.jane.aran, Aug 15, 2023
3. ### eski

Joined:
Jul 31, 2023
Messages:
6
1
This relates to a property in algebra and polynomials and I think this relationship is what you're getting at:

(n^(a+b)) = (1/(n^-(b+a))) IIRC

then you have n^(a) = n^(a+b)/n^b basically and can easily derive any other proofs

#### Attached Files:

• ###### demonstrate.png
File size:
134.8 KB
Views:
6
eski, Aug 16, 2023