Exponential equations

Here's something nontraditional:

2x3x/21=02^x-3^{x/2}-1=0

We can guess or use W|A to figure that x=2x=2. But can we crack this problem analytically?

This problem arose out of Karim Mohamed's solution to Big Problem, small solution.

Note by John Muradeli
5 years, 3 months ago

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Do you mean finding a general solution in terms of functions?(logarithms for example) or just approximating and finding number of solutions????

Hasan Kassim - 5 years, 3 months ago

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Well, general solution and a method of solving... Is there one? Like, you know, ax2+bx+c=0x=b±b24ac2aax^2+bx+c=0 \Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, and so on. Does this form of equations have a general solution? If not, that's ok. Just a solution to 2x3x/21=02^x-3^{x/2}-1=0 would be fine too.

Thanks

John Muradeli - 5 years, 3 months ago

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A non-rigorous proof would be to note how

2x3x/22^x-3^{x/2}

grows larger and larger as xx increases. A quick table would quickly confirm. If we let y=2x3x/2y=2^x-3^{x/2}, we have:

(2,1)(2, 1),

(4,5)(4, 5),

(6,37)(6, 37),

(8,175)(8, 175),

and so on and so forth.

Finn Hulse - 5 years, 3 months ago

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Not seeking for a proof here, but an analytical solution. What if we had instead

2.71828x6.67384x/2360=02.71828^x-6.67384^{x/2}-360=0?

Try making tables now.

John Muradeli - 5 years, 3 months ago

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Can this even be done? Are equations in form ax+bx/2+c=0a^x+b^{x/2}+c=0 solvable? Seems tempting to complete the square, until we realize it doesn't work like that on exponentials.

Oh, maths.

Thanks, Finn.

John Muradeli - 5 years, 3 months ago

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