## Hero Group Title $\large3^{(x+1)}+ 5 = 7^{(x-1)}$ one year ago one year ago

1. akash809 Group Title

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

2. Hero Group Title

lol, what?

3. asnaseer Group Title

Numerical or algebraic solution?

4. asnaseer Group Title

If Numerical then you can use the Newton-Raphson method

5. Hero Group Title

Algebraic

6. asnaseer Group Title

Then things get a lot more interesting...

7. asnaseer Group Title

Do you know how to solve (i.e. is this a challenge)?

8. Hero Group Title

If it was a challenge, I would have posted that it was a challenge.

9. asnaseer Group Title

:) fair enough - no harm in asking though

10. asnaseer Group Title

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....

11. asnaseer Group Title

\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}

12. asnaseer Group Title

I have no idea whether or not this is the right approach - just thinking out aloud

13. Hero Group Title

It was a good effort

14. asnaseer Group Title

@KingGeorge - any thoughts on this?

15. KingGeorge Group Title

Nothing yet, but I'm certainly thinking about it.

16. asnaseer Group Title

another thought...

17. Hero Group Title

Okay, I guess this can be an unofficial challenge then

18. asnaseer Group Title

$7^x-21*3^x-35=0$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?

19. KingGeorge Group Title

$7^x*3^x-21*3^{2x}-35*3^x=-21\cdot3^{2x}+3^x(7^x-35)=0$Perhaps the quadratic formula could be used to solve for $$3^x$$.

20. KingGeorge Group Title

We get $\large 3^x=\frac{(35-7^x)\pm\sqrt{(7^x-35)^2+84}}{-42}$Which certainly doesn't seem very nice.

21. asnaseer Group Title

This equation is just not playing ball with us :(

22. KingGeorge Group Title

It doesn't seem to simplify very nicely either

23. asnaseer Group Title

@Hero can this definitely be solved algebraically?

24. asnaseer Group Title

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.

25. KingGeorge Group Title

I've got to go. I'll check back later.

26. asnaseer Group Title

or maybe there is something in number theory that can be used here because 3 and 7 are co-prime?$7^x-21*3^x-35=0$

27. asnaseer Group Title

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.

28. Hero Group Title

It cannot be solved algebraically, but I was just making sure.