Find Domain Of: √(log1/5(1/Sin5x))

[Plutonium88]

√

Homework Statement

√(log1/5(1/Sin5x))

Log to the base 1/5*(1/Sin5x)


FIND THE DOMAIN


I have an answer to this solution, I Just need some help with understanding it :(... and also could someone tell me if info is missing.


f(x) = √(log1/5(1/Sin5x))

Restrictions:

1/sin5x ǂ 0

sin5x ǂ 0

5x = ∏n
xǂ (∏n)/5

√(log1/5(1/Sin5x)) ≥ 0
(square both side)
(log1/5(1/Sin5x)) ≥ 0

1/sin5x ≤ 1

Sin5x ≥ 1

**** Sin M ≤ 1 ***** (this is refferring to the slope)


Sin5x = Sin(∏/2 + 2∏n)

5x = = ∏/2 + 2∏n
x = ∏/10 + 2∏n/5

ans: XE(∏/10 + 2∏n/5]

anyways i would appreciate if some one could help me understand this solution better. <3

 

[HallsofIvy]

I assume you mean log base 1/5:
\sqrt{log_{1/5}(\frac{1}{sin(5x)})}
Yes, the sin(5x) in the denominator cannot be 0. Sine sin(x) is 0 for x any multiple of \pi, that means 5x cannot be as multiple of \pi:x\ne n\pi/5 for any integer n, as you say.

But also, logarithm is only defined for positive numbers so in addition to sin(5)\ne 0, we must have sin(5x)> 0. sin(\theta) is positive for 0&lt; x\le \pi, 2\pi&lt; x&lt; 3\pi, etc. We can write that as 2n\pi&lt; x&lt; (2n1)\pi for any integer n.

I don't know why you are looking at "\sqrt{log_{1/5}(1/sin(5x)}\ge 0". That is a statement about the value or range of the function and has nothing to do with the domain. Rather, what is important is that in order that a square root exist, the argument must be non-negative. And \log(x)\ge 0 if and only if x\ge 1. So we must have log_{1/5}(1/sin(5x))\ge 1 which means that we must have 1/sin(5x)\ge 1/5[/tex] which, since we have already required that it be positive, is the same as 5\ge sin(5x). For what x is that true?

 

[Plutonium88]

I assume you mean log base 1/5:
\sqrt{log_{1/5}(\frac{1}{sin(5x)})}

Yea my bad D:

Yes, the sin(5x) in the denominator cannot be 0. Sine sin(x) is 0 for x any multiple of \pi, that means 5x cannot be as multiple of \pi:x\ne n\pi/5 for any integer n, as you say.

This is clear.

But also, logarithm is only defined for positive numbers so in addition to sin(5)\ne 0, we must have sin(5x)> 0. sin(\theta) is positive for 0&lt; x\le \pi, 2\pi&lt; x&lt; 3\pi, etc. We can write that as 2n\pi&lt; x&lt; (2n1)\pi for any integer n.

I don't know why you are looking at "\sqrt{log_{1/5}(1/sin(5x)}\ge 0". That is a statement about the value or range of the function and has nothing to do with the domain.

Rather, what is important is that in order that a square root exist, the argument must be non-negative. And \log(x)\ge 0 if and only if x\ge 1. So we must have log_{1/5}(1/sin(5x))\ge 1 which means that we must have 1/sin(5x)\ge 1/5[/tex] which, since we have already required that it be positive, is the same as 5\ge sin(5x). For what x is that true?

<br /> <br /> When <br /> 5≥sin5x <br /> <br /> For all X values, Sin5x will be less than 5..