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Hero

\[\large3^{(x+1)}+ 5 = 7^{(x-1)}\]

  • one year ago
  • one year ago

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  1. akash809
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    awesome stuff..wl answer after half an hour just doing a bit of work.. u did that 3^x.3+5=7^x/7 write

    • one year ago
  2. Hero
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    lol, what?

    • one year ago
  3. asnaseer
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    Numerical or algebraic solution?

    • one year ago
  4. asnaseer
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    If Numerical then you can use the Newton-Raphson method

    • one year ago
  5. Hero
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    Algebraic

    • one year ago
  6. asnaseer
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    Then things get a lot more interesting...

    • one year ago
  7. asnaseer
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    Do you know how to solve (i.e. is this a challenge)?

    • one year ago
  8. Hero
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    If it was a challenge, I would have posted that it was a challenge.

    • one year ago
  9. asnaseer
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    :) fair enough - no harm in asking though

    • one year ago
  10. asnaseer
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    I haven't solved this type of problem algebraically before, but I can think of some things that might trigger some thoughts from the others....

    • one year ago
  11. asnaseer
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    \[\begin{align} 3^{x+1}+5&=7^{x-1}\\ \therefore 3*3^x+5&=7^x/7\\ \therefore 21*3^x+35&=7^x\\ \therefore 7^x-21*3^x-35&=0\\ \therefore (5+2)^x-21(5-2)^x-35&=0 \end{align}\]

    • one year ago
  12. asnaseer
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    I have no idea whether or not this is the right approach - just thinking out aloud

    • one year ago
  13. Hero
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    It was a good effort

    • one year ago
  14. asnaseer
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    @KingGeorge - any thoughts on this?

    • one year ago
  15. KingGeorge
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    Nothing yet, but I'm certainly thinking about it.

    • one year ago
  16. asnaseer
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    another thought...

    • one year ago
  17. Hero
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    Okay, I guess this can be an unofficial challenge then

    • one year ago
  18. asnaseer
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    \[7^x-21*3^x-35=0\]multiply all by \(3^x\) to get:\[7^x*3^x-21*3^{2x}-35*3^x=0\]I wonder if this can be factorised in some way?

    • one year ago
  19. KingGeorge
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    \[7^x*3^x-21*3^{2x}-35*3^x=-21\cdot3^{2x}+3^x(7^x-35)=0\]Perhaps the quadratic formula could be used to solve for \(3^x\).

    • one year ago
  20. KingGeorge
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    We get \[\large 3^x=\frac{(35-7^x)\pm\sqrt{(7^x-35)^2+84}}{-42}\]Which certainly doesn't seem very nice.

    • one year ago
  21. asnaseer
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    This equation is just not playing ball with us :(

    • one year ago
  22. KingGeorge
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    It doesn't seem to simplify very nicely either

    • one year ago
  23. asnaseer
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    @Hero can this definitely be solved algebraically?

    • one year ago
  24. asnaseer
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    the only other thoughts I have is to somehow use the binomial expansion of this:\[(5+2)^x-21(5-2)^x-35=0\]to see if anything interesting shows up.

    • one year ago
  25. KingGeorge
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    I've got to go. I'll check back later.

    • one year ago
  26. asnaseer
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    or maybe there is something in number theory that can be used here because 3 and 7 are co-prime?\[7^x-21*3^x-35=0\]

    • one year ago
  27. asnaseer
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    Yeah - I've got to go now as well - I'll also check back sometime tomorrow to see if anyone else had any ideas on this.

    • one year ago
  28. Hero
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    It cannot be solved algebraically, but I was just making sure.

    • one year ago
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