I know it is a no solution problem, but I was taking a look at it through the use of a limit and wanted to see whether this approach was valid.

 

Given that [math]\lim_{n\rightarrow \infty }\frac{x+1}{x}[/math].

 

We can apply this to 1^x = 2, which can be turned into [math]log_{1}(2) = x[/math].

 

[math]x = \frac{log(2)}{log(1)}[/math]

 

Now, this is undefined. Therefore, we can take the limit by applying the above together.

 

[math]x = \lim_{n\rightarrow \infty }\frac{log(2)}{log(\frac{n+1}{n})}[/math]

 

[math]x = \infty[/math]

 

Is this math wrong? I am assuming some of it is, though checked wolfram: http://www.wolframalpha.com/input/?i=limit+of+x+approaching+infinity+of+log%282%29%2Flog%28%28x%2B1%29%2Fx%29

 

The reason this would be inconsistent is because you could choose any value for the top logarithm and still get the same answer. Just interested in the meaning of it.

Your limit does not converge, so it is not really correct to say x = infinity. You can check the limit of 1^x as x-> infinity, you get 1 and not 2.

 

As you have stated, there is no solution to the equation you start with.

You start with [latex]1^x=2[/latex]. Taking logs gives you xlog(1)=log(2). However log(1)=0, so x=log(2)/0.