# Need some help with this algebra!

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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)

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)

0

reply

Report

#2

(Original post by

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)

**nicevans1**)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)

a will cancel and you can find n

BTW are you sure it wasn't an^x

0

reply

Report

#3

**nicevans1**)

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)

0

reply

Report

#4

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)

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)

0

reply

Ok.

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

4=a3^n 9=a2^n

-5=1^n

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

4=a3^n 9=a2^n

-5=1^n

0

reply

Report

#6

(Original post by

Ok.

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

4=a3^n 9=a2^n

-5=1^n

**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

(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.

0

reply

(Original post by

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.

**davros**)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.

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!!

0

reply

Report

#8

(Original post by

This is last question of my core skills 1... we can do this..

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!!

**nicevans1**)This is last question of my core skills 1... we can do this..

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!!

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

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

0

reply

Report

#9

(Original post by

This is last question of my core skills 1... we can do this..

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!!

Sorry!!

**nicevans1**)This is last question of my core skills 1... we can do this..

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!!

Sorry!!

**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

0

reply

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?

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?

0

reply

Report

#11

(Original post by

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?

**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?

From you can rewrite as

and then substitute into the 2nd equation and rearrange.

0

reply

Report

#12

(Original post by

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

**nicevans1**)Just out of interest someone above said to consider them as simultaneous equations. how would you of done it that way?

You had 2 equations

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

They are simultaneous equations

0

reply

(Original post by

It's basically the same working again.

From you can rewrite as

and then substitute into the 2nd equation and rearrange.

**davros**)It's basically the same working again.

From you can rewrite as

and then substitute into the 2nd equation and rearrange.

0

reply

Report

#14

(Original post by

This is beyond my knowledge... E.g I don't see why the 9 and the a swapped and the 2 became -2...

**nicevans1**)This is beyond my knowledge... E.g I don't see why the 9 and the a swapped and the 2 became -2...

If then .

Now suppose c is something like . Then .

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top