MHB Working on evaluating an 'e' equation to find the domain of an equation

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The discussion focuses on finding the domain of the inequality \(1 \ne e^{1-x^2}\). Participants suggest using the natural logarithm to evaluate the inequality, leading to the transformation \(0 \ne 1 - x^2\). This results in the conclusion that \(x \ne 1\), but further analysis reveals that \(x \ne -1\) must also be considered. The conversation highlights the importance of recognizing both solutions and mentions the possibility of complex solutions as well. The final consensus is that the domain excludes both \(x = 1\) and \(x = -1\).
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Find the domain of the equation

$$1 \ne e^{1-x^2}$$

Is there a way to evaluate this function?
 
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tmt said:
$$1 \ne e^{1-x^2}$$

Is there a way to evaluate this function?

Hint: take $\ln$ on both sides.

Btw, this is not a function nor an equation. It's called an inequality.Moderator's note: I have moved part of your title to your opening post.
Please put all relevant information in your post and do not put part of the question only in the title.
 
tmt said:
Find the domain of the equation

$$1 \ne e^{1-x^2}$$

Is there a way to evaluate this function?

Think about $e^x$. When does it take the value $1$? Can you use this?
 
I like Serena said:
Hint: take $\ln$ on both sides.

Btw, this is not a function nor an equation. It's called an inequality.Moderator's note: I have moved part of your title to your opening post.
Please put all relevant information in your post and do not put part of the question only in the title.

so

$$ln1 \ne ln{e ^{1 - x^2}}$$

which goes to

$$0 \ne 1- x^2$$

which means

$$x \ne 1$$

Is this right?
 
You have missed the case of $x = -1$. Note that $\log$-ing both side (which you have done correctly (Yes)), you get

$$0 \neq x^2 - 1$$

Note that $x^2 - 1$ has TWO solutions : $x = 1$ and $x = -1$. Thus, $x \neq 1$ AND $x \neq -1$

PS : I presume you were asked to find real $x$s? If not, there are also a lot of cases you are missing. For example, if $x = \sqrt{1+2\pi i}$ then $\exp(1-x^2) = 1$.
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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