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This cannot be right... (Problem with Cymath?)

Problem Statement
(1/[2^x])^(1/(2x)) = (2^0.5)/2
Relevant Equations
Exponential proofs.
So basically I decided to simplify the terms on the left, and I got ~0.707. I looked at it and obviously realized it equals sqrt(2)/2. So x = all real #’s.
Since I rarely see problems with infinite solutions, I went on Cymath to confirm it. I understood it all till the very end, where it stated 1/(2^0.5) =/= (2^0.5)/2 .
Here: https://www.cymath.com/answer?q=((1#(2^x)))^(1#(2x))=sqrt(2)#2

Is this an error with the system, because if you rationalize the 1/1.414... , you should get 1.414.../2.
 

BvU

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Which is the same :biggrin:

But: can you do it without crutches ? -- that tell you there's no answer o0)
 

SammyS

Staff Emeritus
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Problem Statement: (1/[2^x])^(1/(2x)) = (2^0.5)/2
Relevant Equations: Exponential proofs.

So basically I decided to simplify the terms on the left, and I got ~0.707. I looked at it and obviously realized it equals sqrt(2)/2. So x = all real #’s.
Since I rarely see problems with infinite solutions, I went on Cymath to confirm it. I understood it all till the very end, where it stated 1/(2^0.5) =/= (2^0.5)/2 .
Here: https://www.cymath.com/answer?q=((1#(2^x)))^(1#(2x))=sqrt(2)#2

Is this an error with the system, because if you rationalize the 1/1.414... , you should get 1.414.../2.
When I first looked at Post #1, I was a bit confused. Were you solving the equation, or were you merely simplifying the left hand side of that equation?

It's fairly apparent that you were to solve the equation, and you were surprised to find that the left hand side simplifies to be equivalent to the right hand side. So ##x## can be any number: well, any number except zero, since ##\dfrac 1 0 ## is undefined.

By the way, the left hand side of the equation is approximately 0.707, not ~0.707 .

As for cymath:
Following the link you provided, we do see something very puzzling for one of the steps cymath takes.
You gave the right hand side to cymath as ##\dfrac{\sqrt 2}{2} ##. That's perfectly fine.

cymath takes the left hand side of the equation and simplifies it. From the 4th step on, here is a snip of what cymath gives.

243681

It's interesting that cymath does not recognize rationalizing the denominator.

By the way, if you input the right hand side as 1/sqrt(2), then cymath gives the correct result.
243682


Another approach to solving this problem is more in line with what you gave in "Relevant Equations", namely Exponential Proofs.

Use logarithms to solve this. I suggest taking ##\log _2 ## of both sides.
.
 
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