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[PTTG??]

2^x=n, solve for x, where n is a power of ten.
Is there more than one answer?

I don't know myself.

[Heron TSG]

There are infinite powers of ten. Therefore, there are infinite answers.

[PTTG??]

There are infinite powers of ten. Therefore, there are infinite answers.

Potentially; but can you prove that there is at least one x where x^2 results in a number that is one followed by any number of zeros and only zeros? Perhaps it would be better to ask what the lowest value of x is.

[SolarShado]

Note: your equation could be better written as "2^x=10^y" and then what values of x will result in y being an integer (whole number)"

however, i'm not sure there is any solution. all powers of 2, when reduced to a list of prime factors will look like {2 2 2 ... 2}, but a power of ten, reduced to the same form will have pairs of {5 2}.

in other words, all powers of 10 will be divisible by 2 and 5, but powers of 2 will only be divisible by 2.

I could be wrong, do you have a number that does fit your requirements? Because

Is there more than one answer?

makes it sound like you do.

[G-Flex]

2^x=n, solve for x, where n is a power of ten.
Is there more than one answer?

I don't know myself.

Is x is some PARTICULAR power of 10, there's only one answer.

If n isn't defined though, and you're just asking if there's more than one solution set for 2^x=n, then yeah, of course there's more than one, since the value of x will change depending on what n is.

[The Mad Engineer]

Note: your equation could be better written as "2^x=10^y" and then what values of x will result in y being an integer (whole number)"

however, i'm not sure there is any solution. all powers of 2, when reduced to a list of prime factors will look like {2 2 2 ... 2}, but a power of ten, reduced to the same form will have pairs of {5 2}.

in other words, all powers of 10 will be divisible by 2 and 5, but powers of 2 will only be divisible by 2.

I could be wrong, do you have a number that does fit your requirements? Because

Is there more than one answer?

makes it sound like you do.

What about zero?
2^0=10^0

[Heron TSG]

X could be infinitely close to zero. (I think.)

[The Mad Engineer]

But then it wouldn't be an integer.

[Heron TSG]

The original post didn't say integer...

[PTTG??]

Just to clarify; I don't even know if there is one. I'm kind of leaning to no, with what solarshado said.

My original thought was if there was a power of 2 that was 1 followed by any number of zeros, but I like the way this is going.

[SolarShado]

"what about zero?" he says. *facepalm* there's that...  good one TME.

Good point Barbarossa, about the integer thing. I assumed we'd be using positive integers, cuz IME ther're the most commonly used.

I suspect you could find fractional values for x and y to make both sides equal to within a very small amount. If it weren't so late i'd be tempted to try and find such values I'm a nerd that way.

[Ampersand]

There are infinitely many solutions assuming that the values for x and y can be fractional values. If you require an integer solution, there is only Zero.

-0.5382 is one such fractional solution

[Heron TSG]

if X can be negative, you can go down to .000000(times infinite)1

[Strife26]

This reminds of the fun I had messing with my new TI-89 Titanium today.

Did you know that calculators don't like it when you tell them to take quadnomials to powers of over a hundred. I like to play with equations like this. I just lack a calculator right now.

[Derakon]

2^x = 10^y

log_10(2^x) = log_10(10^y)

log_10(2^x) = y

Alternatively,

x = log_2(10^y)

Plug in any value of x you like, and you will get a value for y. Or plug in integer values of y, and get the value of x that will generate them. x will probably be irrational, but it will exist for every y you try.