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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
- Posts: 25

Find all real solutions for

I could use detailed help and/or a solution, the sooner the better.

The only work I have is simplifying and proving that 2^x - 1 and x have the same sign.

Thanks!

!nval!d_us3rnam3

*Last edited by !nval!d_us3rnam3 (2018-07-27 03:25:21)*

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**Alg Num Theory****Member**- Registered: 2017-11-24
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There are precisely three solutions:

Unfortunately I do not know how to prove it rigorously. Maybe Bob Bundy can come up with more helpful information.

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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
- Posts: 25

Preferably today.

Can you give me some pointers, at least? How you solved this.

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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
- Posts: 25

Ok, but I wasn't looking for a proof with the answers in the beginning. I've gotten pretty far, but I need to find all the answers to

EDIT: With a rigorous demonstration that this is the answer.

*Last edited by !nval!d_us3rnam3 (2018-07-27 11:02:45)*

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!nval!d_us3rnam3 wrote:

That depends on what you consider to be a 'rigorous demonstration'. Clearly is a factor, since both the LHS and RHS vanish. Dividing your equation by (taking ) gives youSince vanishes whenever , then the numerator of the term on the RHS also vanishes for these values of . That's three solutions, and you can 'rigorously demonstrate' that there aren't any more solutions over via the method in post #4.
Ok, but I wasn't looking for a proof with the answers in the beginning. I've gotten pretty far, but I need to find all the answers to

EDIT: With a rigorous demonstration that this is the answer.

**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

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**Alg Num Theory****Member**- Registered: 2017-11-24
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Let us apply zetafunc’s method in another way. Dividing by 2 and rearranging gives

Now consider the following table:

Thus we see that outside of {0,±1} the equation has no solution as the LHS and RHS have opposite signs. This proves that there are no solutions other than *x* = 0, ±1.

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