How would I solve: 3108416303=a^2010-a^1970?

Question

ganeshie8 · Answer

can u tell us the source of this problem ? school/contest ?

solomonzelman · Answer

looks like a prime, although I wouldn't say for sure.
You can't divide by 3, because the sum of all digits is not divisible by 3,
and it is not an even number, nor it ends with a 5.

thesecret20111 · Answer

I've been working on a question for school but have got stuck here. The calculator can solve this but I'm not sure where to start myself.

thesecret20111 · Answer

If it is indeed a prime what does that mean as far as solving?

solomonzelman · Answer

Yeah, just verified on wolfram, it is a prime.

anonymous · Answer

you can get logarithm  from that equation.

solomonzelman · Answer

No, no use, because the variables are NOT in exponents.

anonymous · Answer

log(c)=2010*log(a) - 1970*log(a)=40*log(a)=log(c)
AM I RIGHT?

solomonzelman · Answer

You mean,    $\normalsize\color{black}{ 3108416303=a^{2010}-a^{1970} }$

$\normalsize\color{black}{ \log 3108416303=\log (a^{2010}-a^{1970}) }$     ?

solomonzelman · Answer

This is even worse than before

solomonzelman · Answer

it is not log(a)-log(b), it is log(a-b)

solomonzelman · Answer

I would leave it when I say that http://www.wolframalpha.com/input/?i=3108416303%3Da%5E2010-a%5E1970.+solve+for+a.

thesecret20111 · Answer

Wait so logs aren't going to help here are they?

solomonzelman · Answer

I don't think so

thesecret20111 · Answer

So should I just wrote down that I used calculator to solve for a?

solomonzelman · Answer

okay, what will you do after this?
you can say it is a product of 3 numbers, as
a²-b²= ....
because a^40= (a^20)²

solomonzelman · Answer

and product of 4 numbers and 5 number

solomonzelman · Answer

It is a quite silly problem, to my (childish) opinion

thesecret20111 · Answer

A silly problem? I was simply asking if this was possible to work out and how I'd go about it.