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

1. akash809

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

lol, what?

3. asnaseer

Numerical or algebraic solution?

4. asnaseer

If Numerical then you can use the Newton-Raphson method

5. Hero

Algebraic

6. asnaseer

Then things get a lot more interesting...

7. asnaseer

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

8. Hero

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

9. asnaseer

:) fair enough - no harm in asking though

10. asnaseer

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

\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

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

13. Hero

It was a good effort

14. asnaseer

@KingGeorge - any thoughts on this?

15. KingGeorge

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

16. asnaseer

another thought...

17. Hero

Okay, I guess this can be an unofficial challenge then

18. asnaseer

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

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

20. KingGeorge

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

This equation is just not playing ball with us :(

22. KingGeorge

It doesn't seem to simplify very nicely either

23. asnaseer

@Hero can this definitely be solved algebraically?

24. asnaseer

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

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

26. asnaseer

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

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

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