Need some help with this algebra!

The equation of a curve is y=ax^n. Given that points (2,9) & (3,4) lie on the curve, calculate the values of a and n?

So this is where i'm at

9 = a2^n and 4 =a3^n

I cannot see how I'm to progress and work out a and n?

Some gently nudging in the right direction might help, (but might not aswell) :colondollar:

Reply 1

TenOfThem

Original post by nicevans1

Divide one equation by the other

a will cancel and you can find n

BTW are you sure it wasn't an^x

Reply 2

tombayes

Original post by nicevans1

divide and conquer or use logs

(edited 9 years ago)

Reply 3

Flame Alchemist

The first step is to consider them as simultaneous equations. Isolate a from one of them and substitute that into the other.

From this derived equation, work out n by taking logs and rearranging. Then, you can simply sub that back in to find a.

(diving one equation by the other will also work; it's simply another method of dealing with simultaneous expressions)

(edited 9 years ago)

Reply 4

nicevans1

Ok.

I thought about the simultaneous route yesterday, but this is what I'm getting!

4=a3^n 9=a2^n

-5=1^n

Reply 5

davros

Original post by nicevans1

Ok.

I thought about the simultaneous route yesterday, but this is what I'm getting!

4=a3^n 9=a2^n

-5=1^n

How exactly have you got that??

(I think I know, but if you've done what I think then you've used incorrect logic to do it!)

As suggested earlier, dividing one equation by the other is the way to start because this will eliminate a and let you focus on finding n first.

Reply 6

nicevans1

Original post by davros

This is last question of my core skills 1... we can do this.. :s-smilie:

4=a3^n 9=a2^n

9 from 4 = -5
a from a = 0
2^n from 3^n = 1^n

when i was writing it, it felt wrong, but i just wanted to try the simultaneous way because Im usually good at them, but to be fair not done any with algerbraic indices before!!

Anyway, the divisional way.
I really cant see what the next step is?
I'm struggling with dividing 3^n by 2^n

Sorry!!

Reply 7

TenOfThem

Original post by nicevans1

This is last question of my core skills 1... we can do this.. :s-smilie:

3^n-2^n is not 1^n

3^n divided by 2^n = (3/2)^n

Reply 8

davros

Original post by nicevans1

This is last question of my core skills 1... we can do this.. :s-smilie:

In addition to what TenOfThem has put, you can't subtract "a from a" because the a's are not added to anything in the first place - they are multiplying something!

To do what you're trying would result in: -5 = a(3^n - 2^n)
and you can't really go any further than that!

Use the hint given by TenOfThem and keep going :smile:

Reply 9

nicevans1

Thanks

I got there.

4/9 = (3/2)^n

4*2^n = 9*3^n

2^(n+2) = 3^(n+2)

so the power must equal 0 and so n must be -2

Just out of interest someone above said to consider them as simultaneous equations. how would you of done it that way?

Reply 10

davros

Original post by nicevans1

It's basically the same working again.

From

9 = a(2^n)

you can rewrite as

a = 9(2^{-n})

and then substitute into the 2nd equation and rearrange.

Reply 11

TenOfThem

Original post by nicevans1

Just out of interest someone above said to consider them as simultaneous equations. how would you of done it that way?

This suggests that you see Simultaneous Equations as a method

You had 2 equations

They needed to be solved at the same time to give the same results

They are simultaneous equations