Question

The variables x and y satisfy the equation
x^{n}y = C, where n and C are constants. When
x = 1.10,y = 5.20, and when x = 3.20, y = 1.05. (i) Find the values of n and C.

Answer 1

\[x^{n}y = C\]

Answer 2

two equation two variables. that should be solvable.

Answer 3

I got to the point \[{1.10^n \over 3.20^n}=0.2019\]

Answer 4

Thats even excellent.

I was gong to say,
take log on both sides,

n ln x + ln y = ln c
put values, you will have two linear equations on n and ln c, solve them and find values.

Answer 5

So how do you get an answer from there?

Answer 6

(1.10/3.20)^n = 2.019

Answer 7

And from there?

Answer 8

take log on both sides.

Answer 9

n = ln(2.019)/ln(1.10/3.20)

Answer 10

n=-0.658

Answer 11

?

Answer 12

could be, now find the value of C using this n.

Answer 13

4.88=c?

Answer 14

does it fit both data??
it it fits, then it's right ... else it's wrong.

Answer 15

OK.

Answer 16

I got 0.488 and 4.88??? weird..

Answer 17

for which value

Answer 18

for C

Answer 19

http://www.wolframalpha.com/input/?i=solve+1.10%5En+*+5.20+%3Dc+%3B+3.20%5En+*1.05+%3D+c

Answer 20

sorry, i mislead you
(1.10/3.20)^n = 2.019

should have been

=> (1.10/3.20)^n = 0.2019

Answer 21

Yes, it works now. Thanks :)