# n degree function

#### furor celtica

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?

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#### Bacterius

Salut,

if $$\displaystyle y = x^n$$ 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 $$\displaystyle x = 2$$. Therefore :

$$\displaystyle 200 < 2^n < 2000$$

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

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

$$\displaystyle 7.64 < n < 10.96$$

So :

$$\displaystyle 7 < n < 11$$

(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 $$\displaystyle \boxed{n = 8}$$, $$\displaystyle \boxed{n = 9}$$ and $$\displaystyle \boxed{n = 10}$$

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

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

(Tongueout)

#### furor celtica

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

#### Bacterius

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 (Worried))

#### furor celtica

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

#### Bacterius

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.

• furor celtica

#### Naim

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!

• furor celtica