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

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

I don't know why you are looking at "[itex]\sqrt{log_{1/5}(1/sin(5x)}\ge 0[/itex]". 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 [itex]\log(x)\ge 0[/itex] if and only if [itex]x\ge 1[/itex]. So we must have [itex]log_{1/5}(1/sin(5x))\ge 1[/itex] which means that we must have [itex]1/sin(5x)\ge 1/5[/tex] which, since we have already required that it be positive, is the same as [itex]5\ge sin(5x)[/itex]. For what x is that true?

 

[Plutonium88]

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

Yea my bad D:

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

This is clear.

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

I don't know why you are looking at "[itex]\sqrt{log_{1/5}(1/sin(5x)}\ge 0[/itex]". 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 [itex]\log(x)\ge 0[/itex] if and only if [itex]x\ge 1[/itex]. So we must have [itex]log_{1/5}(1/sin(5x))\ge 1[/itex] which means that we must have [itex]1/sin(5x)\ge 1/5[/tex] which, since we have already required that it be positive, is the same as [itex]5\ge sin(5x)[/itex]. For what x is that true?

When
5≥sin5x

For all X values, Sin5x will be less than 5..