Alright. On a whim, I started wondering whether there were any answers to the equation 2^x = 10^y for which x and y were both integers. I solved for y, and then wrote a QBASIC program to check every possible integer value of x and then tell me if it finds one that returns a y value.
So far, nothing.
So is it possible to tell whether or not this equation has a solution, and if so, can it tell what the solution is in a better way than guess/check?
Brian
The only solution in the integers is x = y = 0. The reason for this is that 2[sup]x[/sup] will never be a multiple of 5 for x > 0, and 10[sup]y[/sup] will always be a multiple of 5 for y > 0.
Yep.
On the other hand, there are some that are close–which I assume prompted the question. x=10 and y=3 result in 1024=1000, which is close enough that the approximation is used all the time in the computer industry, a kilobyte being 1024 bytes for instance.
Taking the logarithms (base 10) of both sides, leaves
log 2 = y/x
This would be easy if the log of 2 were rational, but the log of 2 is not only irrational, it is transcendental, so there is no chance. However, log 2 = .3010299956, which explains why x=10 and y=3 work so well. Also, 2^100000 is pretty close to 10^30102, relatively. I have 'em as .99900209301438450794403276433e+30103 and 1e+30103.