0
$\begingroup$

is it possible to express $y$ in this expression:

$y^{x} = e^{\sin x}$

through natural logarithm or, maybe, something else, so we get rid of $y$ at all?

I need to have "something in power of $x$" instead of $e^{\sin x}$ , but I am not sure if it is possible at all.

$\endgroup$
11
  • 5
    $\begingroup$ Raise both sides to the $\frac 1x$ power. $\endgroup$ Commented May 1, 2020 at 20:22
  • 2
    $\begingroup$ so, $y = e^{\frac{\sin x}{x}}$ ? $\endgroup$ Commented May 1, 2020 at 20:27
  • $\begingroup$ Not so quick. You need to address domain $\endgroup$ Commented May 1, 2020 at 20:38
  • $\begingroup$ @imranfat what do you mean? $\endgroup$ Commented May 1, 2020 at 20:42
  • 1
    $\begingroup$ In your first equation, x=0 is allowed, in your edit equation, it's not. $\endgroup$ Commented May 1, 2020 at 20:44

1 Answer 1

Reset to default
1
$\begingroup$

The value of $y$ can be computed from the value of $x$ if and only if $x\ne0$. So you can safely raise to $1/x$ both sides.

$$ y=\exp\Bigl(\frac{\sin x}{x}\Bigr) $$

Then this function can be extended by continuity at $x=0$.

If you want to compute $$ \lim_{x\to0}\frac{e^{\sin3x}-e^{\sin x}}{x} $$ do first $$ l_k=\lim_{x\to0}\frac{e^{\sin(kx)}-1}{x} $$ by substituting $y=e^{\sin(kx)}-1$, so you get $$ \lim_{y\to0}\frac{ky}{\arcsin\log(1+y)}= \lim_{y\to0}\frac{ky}{\log(1+y)}\frac{\log(1+y)}{\arcsin\log(1+y)}=k $$ Then your limit is $l_3-l_1=3-1=2$.

$\endgroup$
1
  • $\begingroup$ Yes, that's a much better way of putting it. $\endgroup$ Commented May 1, 2020 at 20:51

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.