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Tricky exponent problem.

Silvanoshei

New member
Nov 17, 2013
5
I'm trying to manipulate (x+1)^x+1 / ((x+1)+1)^x+1

So that I have a 1 in the numerator. If I bring the numerator down using the integer exponent rule, I'll have...
1 / ( (x+1) / (x+1) + 1 )^x+1 ?

Whoops, that's not right...

1 / (x+1)^x * ((x+1)+1)^x+1 ?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I'm trying to manipulate (x+1)^x+1 / ((x+1)+1)^x+1

So that I have a 1 in the numerator. If I bring the numerator down using the integer exponent rule, I'll have...
1 / ( (x+1) / (x+1) + 1 )^x+1 ?

Whoops, that's not right...

1 / (x+1)^x * ((x+1)+1)^x+1 ?
Hi Silvanoshei, welcome to MHB! :)

I'm gambling on what you intend, since I suspect there are a couple of parentheses missing.
So I'll write in $\LaTeX$ what I think you intended.

$$\frac {(x+1)^x+1} {((x+1)+1)^x+1} \overset{?}{=} \frac 1 {(x+1)^x * ((x+1)+1)^x+1}$$

Can you confirm this is what you intended?
Or perhaps show what you did intend?
Either way, this equality does not hold.
 

Silvanoshei

New member
Nov 17, 2013
5
Yeah, on top it's (x+1)^(x+1)
Bottom is ((x+1)+1)^(x+1)
*sorry for the confusion.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Yeah, on top it's (x+1)^(x+1)
Bottom is ((x+1)+1)^(x+1)
*sorry for the confusion.
Like this?
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} \overset{?}{=} \frac 1 {(x+1)^x * ((x+1)+1)^{x+1}}$$
It still makes little sense to me.


What you can do for instance, is:
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} = \left( \frac {x+1} {(x+1)+1}\right)^{x+1}$$
 

Silvanoshei

New member
Nov 17, 2013
5
Like this?
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} \overset{?}{=} \frac 1 {(x+1)^x * ((x+1)+1)^{x+1}}$$
It still makes little sense to me.


What you can do for instance, is:
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} = \left( \frac {x+1} {(x+1)+1}\right)^{x+1}$$
Yup! That's the problem, x is going to go to infinity, so I was trying to manipulate it 1/something to make it zero. If the exponent is x+1 over the fraction, it'll just go to infinity.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Yup! That's the problem, x is going to go to infinity, so I was trying to manipulate it 1/something to make it zero. If the exponent is x+1 over the fraction, it'll just go to infinity.
Continuing:
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}}
= \frac {1} {\left( 1+\frac 1 {x+1}\right)^{x+1}}
$$

If we replace x+1 by n, which also goes to infinity, we get:
$$\frac {1} {\left( 1+\frac 1 n\right)^n}$$

The denominator is a known limit, which approaches $e$ when $n \to \infty$.

So:
$$\lim_{x \to \infty} \frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} = \frac 1 e$$
 

Silvanoshei

New member
Nov 17, 2013
5
Continuing:
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}}
= \frac {1} {\left( 1+\frac 1 {x+1}\right)^{x+1}}
$$
How does this step work?? Not getting it. :(
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
How does this step work?? Not getting it. :(
Let me write out a couple more intermediary steps...
\begin{aligned}
\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}}
&= \left( \frac {x+1} {(x+1)+1}\right)^{x+1} \\
&= \left( \frac {1} {\frac{(x+1)}{x+1}+\frac 1 {x+1}}\right)^{x+1} \\
&= \left( \frac {1} {1+ \frac 1 {x+1}}\right)^{x+1} \\
&= \frac {1} {\left( 1+ \frac 1 {x+1}\right)^{x+1}}
\end{aligned}
 

Silvanoshei

New member
Nov 17, 2013
5
Thanks! I was trying to multiply top and bottom by 1/(x+1)^(x+1), but putting it in fractional over everything form is key! (Heidy)