1^x = 2, is this a correct way to look at it?

[Unity+]

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.

[ajb]

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.

[mathematic]

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.