Thread: Nonlinear Algebra

I don't know if this part of the forum is the appropriate place for this kind of a question but I think you guys can still help solve my problem.
My Chemistry teacher asked me to find at least two values of x for the expression/equation rather
2^x=4x without using a calculator or without relying much on a calculator.

So I used common sense and say that x=4 and I got stuck from there.

This is what I did: 2^x=2^2(x).By just looking at it, I concluded that x must be 4 for the two sides to be equal. From here, I don't know where to go. Is there a formula(good formula maybe) to solve this kind of a question? What can I do to find the other values of x?
Your help will be highly appreciated.

Originally Posted by McCoy

I don't know if this part of the forum is the appropriate place for this kind of a question but I think you guys can still help solve my problem.
My Chemistry teacher asked me to find at least two values of x for the expression/equation rather
2^x=4x without using a calculator or without relying much on a calculator.

So I used common sense to say that x=4 and I got stuck from there.

This is what I did: 2^x=2^2(x).By just loking at it, I concluded that x must be 4 for the two sides to be equal. From here, I don't know where to go. Is there a formula(good formula maybe) to solve this kind of a question? What can I do to find the other values of x?
Your help will be highly appreciated.

If he expects you to find it without mechanical assistence then he is having you on. The other solution is between 0 and 1 as you can see by looking at at and . Linear interpolation (by eye since the values are so simple) indicates that the solution is near . But I think that this is a transcendental number.

RonL

Originally Posted by CaptainBlack

If he expects you to find it without mechanical assistence then he is having you on. The other solution is between 0 and 1 as you can see by looking at at and . Linear interpolation (by eye since the values are so simple) indicates that the solution is near . But I think that this is a transcendental number.

RonL

Thanks a lot.I will report the two solutions to him and ask him to solve the rest without using any mechanical aid hehehe. Just before, I go ...why do you think 1/3 is a transcendental number?


Thanks once more.

Originally Posted by McCoy

Thanks a lot.I will report the two solutions to him and ask him to solve the rest without using any mechanical aid hehehe. Just before, I go ...why do you think 1/3 is a transcendental number?


Thanks once more.

It's not 1/3 that I think transcendental but the solution near it at approx 0.3099..

When I calculate this to 15 significant figures and feed it into the inverse symbolic calculator it reports no hits.

(also mixed transcendental/algebraic equations charateristicly have transcendental roots, only under special conditions are the roots integer, rational or algebraic numbers).

RonL