1. n degree function

the diagram shows the graph of y=x^n, where n is an integer. given that the curve passes between the points (2,200) and (2, 2000), determine the value of n.
i just dont get this question. what does it mean precisely to 'pass between the points', how is this interpreted algebraically? its just too random for me, any helpful hints?

2. Salut,

if $y = x^n$ and the curve passes between A(2,200) and B(2, 2000), first note that point B is higher than point A. So it means that the curve goes under B, and over A, at $x = 2$. Therefore :

$200 < 2^n < 2000$

You can then take the base-2 logarithm of each member of the inequality to extract $n$ :

$\log_2{(200)} < \log_2{(2^n)} < \log_2{(2000)}$

$7.64 < n < 10.96$

So :

$7 < n < 11$

(by rounding, and noting that the question says that the curve passes between the points, not through the points).

Therefore the only options left are $\boxed{n = 8}$, $\boxed{n = 9}$ and $\boxed{n = 10}$

All those values of $n$ satisfy your statement. So how do we know which one is the good one ?

(!) Look at the curve : the values are negative with $x < 0$, and positive $x > 0$. This is only possible if the exponent is odd (if it is even, the result is always positive). Therefore, the only solution is ... $\boxed{n = 9}$ !

3. merci
dis t'es pas vraiment né en 1993 non?
rassure-moi, sinon merde j'ai la honte

4. Originally Posted by furor celtica
merci
dis t'es pas vraiment né en 1993 non?
rassure-moi, sinon merde j'ai la honte
Désolé j'ai seize ans et demi ^^ (donc 1993 )

5. oh la la
je me sens nul là tout a coup
sinon
is there any other way to get this result sans utiliser de log? parce que this is a revision question for a section in which logarithms hadnt yet been introduced. donc voilà

6. I don't think so. Since the original question basically boils down to solve for an exponent, logs have to be used. Well ... so I think.

7. Same question

Hello there! I had the same question and solved it in a different way. Actually I wasn't sure if it is correct, but I got to the possibilities of n=8, n=9, n=10, and then I saw this thread.
So, when you get to 200<2^n<2000
you can write 25*2^3<2^n<125*2^4
from that you have 2^4*2^3<2^n<2^7*2^4
from which we get 2^7<2^n<2^11
so 7<n<11. From here you get the options that n=8,9,10 (as n is an integer), and from this as 9 is the only odd one, you find the solution. I know this might be a bit late, but I just came across it!