# Need some help with this algebra!

Watch
Announcements
#1
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
5 years ago
#2
(Original post by 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)
Divide one equation by the other

a will cancel and you can find n

BTW are you sure it wasn't an^x
0
5 years ago
#3
(Original post by 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)
divide and conquer or use logs
0
5 years ago
#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)
0
#5
Ok.

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

4=a3^n 9=a2^n

-5=1^n
0
5 years ago
#6
(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.
0
#7
(Original post by 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.
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!!
0
5 years ago
#8
(Original post by 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
5 years ago
#9
(Original post by 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!!
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
0
#10
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?
0
5 years ago
#11
(Original post by 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?
It's basically the same working again.

From you can rewrite as

and then substitute into the 2nd equation and rearrange.
0
5 years ago
#12
(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

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

They are simultaneous equations
0
#13
(Original post by davros)
It's basically the same working again.

From you can rewrite as

and then substitute into the 2nd equation and rearrange.
This is beyond my knowledge... E.g I don't see why the 9 and the a swapped and the 2 became -2...
0
5 years ago
#14
(Original post by nicevans1)
This is beyond my knowledge... E.g I don't see why the 9 and the a swapped and the 2 became -2...
This is GCSE knowledge - have you not covered indices before?

If then .

Now suppose c is something like . Then .
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (94)
13.91%
I'm not sure (32)
4.73%
No, I'm going to stick it out for now (213)
31.51%
I have already dropped out (15)
2.22%
I'm not a current university student (322)
47.63%