No, this is not homework.

How do you solve something like this algebraically:

2^(x^2)-2^(x)= 448

So obviously you could make x=3

2^9 - 2^6 = 512-64 = 448

How do you solve this otherwise?

No, this is not homework.

How do you solve something like this algebraically:

2^(x^2)-2^(x)= 448

So obviously you could make x=3

2^9 - 2^6 = 512-64 = 448

How do you solve this otherwise?

Numerically. There may be a solution involving special functions, but you need numerical techniques to evaluate those anyway, so why not cut out the middle step?

(Also, there’s a mistake in your arithmetic or a typo in the equation. 3 is not a root of what you originally wrote.)

If you can’t solve it, nitpick the question.

If x = 3, the second term shouldn’t be 2^6.

Perhaps you meant your equation to read:

2^(x^2) - 2^(2x) = 448?

I tried logarithms and substitutions, but seemed to always hit dead ends.

Yes, sorry I mistyped originally. You are correct. It was supposed to read:

2^(x^2) - 2^(2x) = 448?

There are two solutions, 3 and about -2.96773. As **ultrafilter** said, numerical evaluation is really your only option in this case. Personally, I just entered it in Wolfram|Alpha.

It should be noted that the vast majority (by which I mean, “all but an infinitesimally small proportion of them”) of math problems you can write down can’t be solved analytically, even with recourse to special functions. Fortunately, many of the math problems people are actually interested happen to fall into that infinitesimally small subset, and many more are very similar to each other, making it practical to invent special functions to deal with them.

I see. So would problem like this still fall into that category.

25^(2x) = 5^(x^2) -12

My first inclination was to re-write the left side as (5^2)^(2x) or 5^(4x). Then move things around such that the problem would read:

5^(x^2) - 5^(4x) = 12

Then I got stuck. I can see that one answer is really close to 4. And of course I can plug it into Wolfram, but is there really no way to do this by hand?

Iterated exponentiation is really uncommon, so it’s unlikely that anyone has ever invented a special function that would cover that one. Sometimes it’s possible to stumble upon a method of doing a very specific problem by hand, but it’s not possible to know in general whether it can be done for any given problem (I’m pretty sure that it’s equivalent to the halting problem). In other words, if you have a problem for which you haven’t found a solution yet, maybe you’ll find it if you just work at it for a little bit longer, but maybe you’ll literally never find it.

By the way, did these problems actually come up in some context?

it can’t be substituted. the “difference” rule only applies to the difference of two separate logs, not the difference of two things within the same log.

If want to try and find an integer solution, work in binary and note that the difference of two powers of two will be a binary number consisting of a block of 1’s followed by a block of zeros.

448 in binary is 111000000 (block of 3 1’s followed by 6 zeros)

= 1000000000 - 1000000

back to decimal, that equals 2^9 - 2^6

x^2=9 and 2x=6 gives the consistent solution x=3 (if not consistent, then there is not an integer solution).

If you take an expression of the form of:

A[sup]x[/sup] - A[sup]-x[/sup] = c

you can express x in terms of c as:

x = log[sub]A[/sub](c/2 + sqrt(c[sup]2[/sup]/4 + 1))

so sometimes there is a nice little trick you can do to get a ‘nice’ answer

But with an expression of the form of

A[sup]x[sup]2[/sup][/sup] - A[sup]2x[/sup] = c

I think you’re out of luck

This is really clever! I think it’s the closest you will come to an analytical solution. Very smart.

The latter one was in a calculus book. The first was just an simplified example of the issue I was coming up against. Thanks to everyone who responded.

Let me guess, in the chapter on Newton’s Method? When you can’t get an analytic answer, you go for a numerical one, and Newton’s Method is a relatively decent method for doing that: It’s simple enough that it can be done by hand, and it’s not too horribly inefficient (in fact, something resembling Newton’s Method may well be how Wolfram got the answers that it did). Plus, it’s also a good illustration of the practical applications of calculus.

Variants of Newton’s method are pretty much state of the art for optimization.