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inequation with just 3 solutions

Vali

Member
Dec 29, 2018
48
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

Administrator
Staff member
Feb 24, 2012
13,736
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
Dec 29, 2018
48
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
Aug 30, 2012
1,121
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.
\(\displaystyle log_3 \left ( \dfrac{n}{x} \right ) \geq 1\)

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
Dec 29, 2018
48
Thank you very much for the help :)