Question 1133315
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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