as per title really! anyone know?
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Maths question - what x such that 2^x = 10y where x,y are integers
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Posted 2 years ago #
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none
Posted 2 years ago # -
10 = 2 * 5 and there are no integer multiples of 2 that make 5
Posted 2 years ago # -
42?
Posted 2 years ago # -
What simon said. 2^x is always going to end up finshing in 2,4,6 or 8, never 0
Posted 2 years ago # -
not quite sure what SFB is getting at, but that equation cant be solved, you need the same number of equations as you have unknowns
eg
if x=1 then
2=10y so
y=0.2
if x=2 then
4=10y so
y= 0.4
if x=3 then
8=10y so
y=0.8
and so on
Posted 2 years ago # -
oops, missed the interger bit
Posted 2 years ago # -
it's 2^5 not 2*5...
i don't know whether it can be done, as 2^x as mentioned, always ends in 2,4,6 or 8 doesn't it?
Posted 2 years ago # -
It doesn't even work if x=0...
So nope, no integer solutions...
Posted 2 years ago # -
Don't worry thisisnotaspoon, I thought I had a working answer but had pressed the wrong minus button on windows calc
Posted 2 years ago # -
and likewise, you wouldn't be able to solve it without another equation, you'd need to go on trial and error.
Posted 2 years ago # -
not quite sure what SFB is getting at, but that equation cant be solved, you need the same number of equations as you have unknowns
the question is asking for one or more solutions, not just one, but in fact there are none. If the 10 were replaced with 8 then there would be an infinite number of them starting at x = 3,4,5 ... and y = 1,2,4 ...
ie x > 2 and y = 2^(x-3)
Posted 2 years ago # -
As far as I can work out SoB is correct.
Posted 2 years ago # -
it's 2^5 not 2*5...
i don't know whether it can be done, as 2^x as mentioned, always ends in 2,4,6 or 8 doesn't it?
I was talking about the 10 in 10y. 10 is 2*5, so any multiple of it will still have 5 as a factor, but no integer multiple of 2 will make 5 (which is prime) so there are no solutions. Actually saying all powers of 2 end in 2,4,6 or 8 amounts to the same thing, as all decimal numbers have to end in a digit and zero doesn't work because all numbers ending in zero are multiples of 5 (ignoring 0 which is not a finite power of 2)
Posted 2 years ago # -
you wouldn't be able to solve it without another equation, you'd need to go on trial and error
I'd question that, the process might be called analysis
divide through by 2 to get 2^(x-1) = 5y,
observe all the possible answers on the left only have 2 as factors and 5 is prime
end ofPosted 2 years ago # -
Simon: Since when has 2^(x-1) = (2^x)/2 ?
Would an easier approach be to assume both x and x must be negative, and ask when is log10(2^x) = integer ?
Saying that, I got no integers out by the time Excel packed up at log (2^-1022)Posted 2 years ago # -
Simon: Since when has 2^(x-1) = (2^x)/2 ?
Stick some values in and see what you come back with
Posted 2 years ago # -
Simon: Since when has 2^(x-1) = (2^x)/2 ?
Since, ummm, always.
assume both x and x must be negative, and ask when is log10(2^x) = integer ?
Assuming you meant "assume both x and y must be negative" this makes no sense.
x is negative => 2^x is positive
y is negative => 10y is negativeTherefore, 2^x = 10y has no solution for negative x and y.
Posted 2 years ago # -
Just graph it you'll soon see the solution then.
Posted 2 years ago # -
Things happen when the mind's on the real job!
Posted 2 years ago #
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Now, if imaginary values of x are permitted...
Posted 2 years ago # -
Simon: Since when has 2^(x-1) = (2^x)/2 ?
it's interesting to speculate if it held true before the concept of number, or of powers, was conceived...
Would an easier approach be to assume both x and x must be negative, and ask when is log10(2^x) = integer ?
Saying that, I got no integers out by the time Excel packed up at log (2^-1022)no real power of 2 whether positive or negative will yield a negative result, and all the negative powers will give a non-integer value less than one. But IMO actually looking at what is happening is more productive than a fishing expedition...
Posted 2 years ago # -
note that x^2 + y^2 = z^2 has an infinite number of integer solutions (right angle triangles) but no one has ever found one for x^n + y^n = z^n for n>2, and Fermat claimed to have an easy proof, though I believe it was only satisfactory reproved recently! Why not try that in Excel ??
Posted 2 years ago # -
Hmmm... Fermat's Last Theorem... that should be easy, after all, I have a book around here that tells me how to do it...
What could be so difficult?
Posted 2 years ago # -
I would say that Ian Munro has given the most sensible and intuitive answer so far. Any one who can multiple a number by 2 and get the correct result will see the recurring pattern that he is referring to within 30 seconds. Or at least this is true for positive values of x.
Posted 2 years ago # -
I would say that Ian Munro has given the most sensible and intuitive answer so far. Any one who can multiple a number by 2 and get the correct result will see the recurring pattern that he is referring to within 30 seconds. Or at least this is true for positive values of x.
I can't find the post this refers to
Posted 2 years ago #
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