Here's something nontraditional:

\(2^x-3^{x/2}-1=0\)

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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Thanks

Log in to reply

A non-rigorous proof would be to note how

\[2^x-3^{x/2}\]

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

\((2, 1)\),

\((4, 5)\),

\((6, 37)\),

\((8, 175)\),

and so on and so forth.

Log in to reply

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

Oh, maths.

Thanks, Finn.

Log in to reply

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

\(2.71828^x-6.67384^{x/2}-360=0\)?

Try making tables now.

Log in to reply