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

**MRich520** · September 23rd 2011, 03:35 PM

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?

**Plato** · September 23rd 2011, 03:58 PM

Originally Posted by MRich520

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

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

Now do you agree that $(\sqrt{e})^0=1~?$
If you do then what does $\ln(\sqrt{e})=~?$

**MRich520** · September 23rd 2011, 04:30 PM

Originally Posted by Plato

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

Now do you agree that $(\sqrt{e})^0=1~?$
If you do then what does $\ln(\sqrt{e})=~?$

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

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

**Plato** · September 23rd 2011, 04:54 PM

Originally Posted by MRich520

$\ln(\sqrt{e})=1/2$
Eh...I'm not sure if that was the answer you were asking for. Mind clarifying?

Well what is $\ln(\sqrt{e})-0.5~?$
And what is $\sqrt{e}$ to that power?

**Soroban** · September 23rd 2011, 07:17 PM

Hello, MRich520!

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

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

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

And we have two equations to solve:

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

**MRich520** · September 28th 2011, 01:35 PM

Originally Posted by Soroban

Hello, MRich520!

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

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

And we have two equations to solve:

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

Thanks!

I should know that from my classes

.

And sorry Plato I forgot to respond

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