How to express y in $y^{x} = e^{\sin x}$ through natural logarithm?

Question

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.

In your first equation, x=0 is allowed, in your edit equation, it's not. — imranfat, May 01 '20 at 20:44
@imranfat If the thing should define $y$ as a function of $x$, certainly $x=0$ is not allowed. After the fact, the function can be extended by continuity at $x=0$. — egreg, May 01 '20 at 20:47
@egreg. True, but the OP did not allude to any of that. I think it is important. — imranfat, May 01 '20 at 20:49
@imranfat yes, I see this problem now. Actually, I am trying to evaluate $\lim_{x \to 0}\frac{e^{\sin 3x} - e^{\sin x}}{x}$ and was hoping that this transformation can help. Now I'm not so sure... — Xentai, May 01 '20 at 20:49
@Xentai That's a simple derivative, isn't it? $3e^{\sin x}\cos3x-e^{\sin x}\cos x$ evaluated at $0$ yields $3-1=2$. — egreg, May 01 '20 at 20:52
@egreg I was trying to find the limit using just knowledge that $\lim_{x\to0}\frac{a^{x}-1}{x}=\ln(a)$, without using derivatives. — Xentai, May 01 '20 at 21:00
@Xentai I see; not really the best way, but it can be done. See edit to my answer. — egreg, May 01 '20 at 21:31

egreg · Accepted Answer · 2020-05-01T21:30:29.630

1

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 $l_3-l_1=3-1=2$.

edited May 01 '20 at 21:30

answered May 01 '20 at 20:50

egreg

238,574

Yes, that's a much better way of putting it. – imranfat May 01 '20 at 20:51

How to express y in $y^{x} = e^{\sin x}$ through natural logarithm?

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