please try, how is it?

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

  1. hall

    hall

    Joined:
    Aug 15, 2023
    Messages:
    1
    Likes Received:
    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!
    please try, how is it?
     
    hall, Aug 15, 2023
    #1
  2. hall

    e.jane.aran

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

    ===============================
    upload_2023-8-15_9-30-51.gif

    Then:

    upload_2023-8-15_9-31-52.gif

    Then:

    upload_2023-8-15_9-32-30.gif = ??

    upload_2023-8-15_9-33-48.gif = ??

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

    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:

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

    eski

    Joined:
    Jul 31, 2023
    Messages:
    6
    Likes Received:
    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:

    eski, Aug 16, 2023
    #3
Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Similar Threads
There are no similar threads yet.
Loading...