zeros

**DINOCALC09** · March 13th 2008, 08:15 AM

How many zeros does the function y = sin(lnx) have for 0<x<1 ?

**ThePerfectHacker** · March 13th 2008, 08:26 AM

Originally Posted by DINOCALC09

How many zeros does the function y = sin(lnx) have for 0<x<1 ?

$\sin (\ln x) = 0 \implies \ln x = \pi n \mbox{ where }n\in \mathbb{Z}$ . Thus, $x = e^{\pi n}$ if $n=-1,-2,-3,...$ then $0 < x < 1$ and it is a zero. Thus, there are infinitely many zeros.

**Moo** · March 13th 2008, 08:26 AM

Hello,

You know that if 0<x<1, ln(x)<0

But within $]- \infty;0[$ there is an infinity of solutions making sin(ln(x))=0, because of the 2pi period.

**colby2152** · March 13th 2008, 08:31 AM

Originally Posted by DINOCALC09

How many zeros does the function y = sin(lnx) have for 0<x<1 ?

$\sin(\ln|x|)=0$

$\ln|x| = \sinh(0) \Rightarrow \ldots -\pi, 0, \pi \ldots$

$x = e^{\pi n}$ for $n \ge 0$ where n is an integer and $0 < x < 1$

Therefore $\pi n < 0$ , so $n < 0$

$x = e^{\pi n} \forall n < 0, n \in Z$

$\forall$ : for all* (isn't this supposed to have a horizontal line through the middle?)

However....

**Moo** · March 13th 2008, 08:32 AM

Write "\forall" in the math tags :-)

**colby2152** · March 13th 2008, 08:34 AM

Originally Posted by Moo

Write "\forall" in the math tags :-)

Thanks!

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