MHB Inequation with just 3 solutions

[Vali]

I have the following function:
f: (0,infinity) -> R
f(x)=log_3(x) (the base is 3)
I need to find the number of "n" values ( n is a natural number except 0 N*) such that this inequation: f(n/x) >= 1 to have just 3 solutions in N*.
A. infinity
B. 6
C. 9
D. 26
E. 3 (correct answer)

 

[MarkFL]

Maybe I'm reading this wrong, but it looks to me like you need to count the number of natural numbers greater than \(3x\), which for any choice of \(x\) allowed, would be countably infinite.

 

[Vali]

Yes, I got the same result n>=3x
Maybe I wrote the sentence is a wrong way because I translated it from romanian.
I posted o picture below.
Exercise number 45.
The number of n values ( n natural ) for which the inequation f(n/x) >= 1 has exactly 3 solutions in N* is: ...

 

[topsquark]

I have the following function:
f: (0,infinity) -> R
f(x)=log_3(x) (the base is 3)
I need to find the number of "n" values ( n is a natural number except 0 N*) such that this inequation: f(n/x) >= 1 to have just 3 solutions in N*.
A. infinity
B. 6
C. 9
D. 26
E. 3 (correct answer)

This is a weird one. Here's my guess.
[math]log_3 \left ( \dfrac{n}{x} \right ) \geq 1[/math]

Since the log function is continuous we can take the exponent of base 3 on both sides:
[math]3^{ log_3 (n/x) } \geq 3^1[/math]

[math]\dfrac{n}{x} \geq 3[/math]

[math]n \geq 3x[/math]

So to have only 3 solutions, n = 3, gives us a domain for x as {1, 2, 3}.

-Dan

 

[Vali]

Thank you very much for the help :)