A variable raised to a power that contains that variable, all equal to one?

I put the following problem into Wolfram Alpha and found that the solutions are the square root of e and 1. Can somebody explain how you'd solve it to get both of those answers? I only got the square root of e. Here's the problem:

x^(ln(x) - 0.5) = 1

It makes sense to me that 1 works as well, but am I missing some kind of rule that would have made me get 1 when I solved it the first time around?

Re: A variable raised to a power that contains that variable, all equal to one?

Quote:

Originally Posted by

**MRich520** x^(ln(x) - 0.5) = 1

I am glad that $\displaystyle x=1$ works for you.

Now do you agree that $\displaystyle (\sqrt{e})^0=1~?$

If you do then what does $\displaystyle \ln(\sqrt{e})=~?$

Re: A variable raised to a power that contains that variable, all equal to one?

Quote:

Originally Posted by

**Plato** I am glad that $\displaystyle x=1$ works for you.

Now do you agree that $\displaystyle (\sqrt{e})^0=1~?$

If you do then what does $\displaystyle \ln(\sqrt{e})=~?$

$\displaystyle \ln(\sqrt{e})=1/2$

Eh...I'm not sure if that was the answer you were asking for. Mind clarifying?

Re: A variable raised to a power that contains that variable, all equal to one?

Quote:

Originally Posted by

**MRich520** $\displaystyle \ln(\sqrt{e})=1/2$

Eh...I'm not sure if that was the answer you were asking for. Mind clarifying?

Well what is $\displaystyle \ln(\sqrt{e})-0.5~?$

And what is $\displaystyle \sqrt{e}$ to that power?

Re: A variable raised to a power that contains that variable, all equal to one?

Hello, MRich520!

Quote:

$\displaystyle \text{Solve for }x\!:\;\;x^{(\ln x - \frac{1}{2})} \:=\: 1$

Take logs: .$\displaystyle \ln\left(x^{(\ln x - \frac{1}{2})}\right) \:=\:\ln 1$

We have: .$\displaystyle \left(\ln x - \tfrac{1}{2}\right)\cdot \ln x \:=\:0$

And we have two equations to solve:

. . $\displaystyle \begin{array}{ccccccc}\ln x - \tfrac{1}{2} \:=\:0 & \Rightarrow & \ln x \:=\:\tfrac{1}{2} & \Rightarrow & x \:=\:e^{\frac{1}{2}} \\ \\ \ln x \:=\:0 & \Rightarrow & x \:=\:e^0 & \Rightarrow & x \:=\:1 \end{array}$

Re: A variable raised to a power that contains that variable, all equal to one?

Quote:

Originally Posted by

**Soroban** Hello, MRich520!

Take logs: .$\displaystyle \ln\left(x^{(\ln x - \frac{1}{2})}\right) \:=\:\ln 1$

We have: .$\displaystyle \left(\ln x - \tfrac{1}{2}\right)\cdot \ln x \:=\:0$

And we have two equations to solve:

. . $\displaystyle \begin{array}{ccccccc}\ln x - \tfrac{1}{2} \:=\:0 & \Rightarrow & \ln x \:=\:\tfrac{1}{2} & \Rightarrow & x \:=\:e^{\frac{1}{2}} \\ \\ \ln x \:=\:0 & \Rightarrow & x \:=\:e^0 & \Rightarrow & x \:=\:1 \end{array}$

Thanks!

I should know that from my classes :p.

And sorry Plato I forgot to respond :(