# inequation with just 3 solutions

#### Vali

##### Member
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

#### MarkFL

##### Pessimist Singularitarian
Staff member
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

##### Member
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: ...

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#### topsquark

##### Well-known member
MHB Math Helper
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
This is a weird one. Here's my guess.
$$\displaystyle log_3 \left ( \dfrac{n}{x} \right ) \geq 1$$

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

$$\displaystyle \dfrac{n}{x} \geq 3$$

$$\displaystyle n \geq 3x$$

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

-Dan

#### Vali

##### Member
Thank you very much for the help