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

[tmt1]

Find the domain of the equation

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

Is there a way to evaluate this function?

 

[I like Serena]

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

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.

 

[mathbalarka]

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?

 

[tmt1]

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?

 

[mathbalarka]

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 $\exp(1-x^2) = 1$.