SOLUTION: Hi need help on this question
The variables x and y are connected by the equation y= ax^n where a and n are constants. Given that y=6360 when x = 10, find the values of a and n
Algebra.Com
Question 1133315: Hi need help on this question
The variables x and y are connected by the equation y= ax^n where a and n are constants. Given that y=6360 when x = 10, find the values of a and n to 3 s.f
this is what I have done lg y = lg a + nlg x
Lg 6360 = lg a +nlg 10
Don't know what to do next. Thanks
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
unless i'm missing something, it appears that a and n can be anything.
if you pick an a, you can solve for n.
if you pick an n, you can solve for a.
for example:
your equation is y = a * x^n
you are given that y = 6360 when x = 10.
your equation becomes 6360 = a * 10^n
you can pick any positive value for a and solve for n.
you can pick any positive or negative value for n and solve for a.
for example:
let a = 5.
the equation becomes 6360 = 5 * 10^n
divide both sides by 5 to get 6360/5 = 10^n
take log of both sides to get log(6360/5) = log(10^n)
this becomes log(6360/5) = n * log(10)
solve for n to get n = log(6360/5)/log(10) = 3.104407111
your equation becomes 6360 = 5 * 10^3.104407111
evaluate to get 6360 = 6360
a can't be negative because they you'll be taking the log of a negative number which is not allowed.
allow n to be equal -5.
the equation becomes 6360 = a * 10^-5
divide by 10^-5 to get 6360 / 10^-5 = a
solve for a to get a = 636000000
equation becomes 6360 = 636000000 * 10^-5
evaluate to get 6360 = 6360
there are an infinite number of values of a and n that will satisfy this equation as long as a is not negative.
the problem in getting a unique solution is that you have one equation in 2 unknowns.
in order to get a unique solution, you need to have the same number of unknowns as the number of equations.
i built a table in excel with various values of n given to show you that there is no limit to the possible values of a from that.
i chose values of n from -10 to +10.
the values of a were calculated, and then the values of a and n were used to solve the equation of y = a*10^n
the value of n can be anything and you will get a value of a that makes the equation true.
that's my thinking.
perhaps the questions was not posed correctly?
when given the value of n, i didn't even need logs.
if given a when a is positive, i used logs to solve for n.
here's the results of excel.
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