You are Here: Home >< Maths

# Logs Logs Logs Watch

1. Can anyone explain how this question is done:

The points (2, 6) and (3, 18) lie on the curve y=ax^n
Use logarithms to find the values of a and n, giving your answers correct to 2 decimal places.

Many thanks in advance
2. Have you tried forming a system of two simultaneous equations in terms of a and n?

(Hint: The points (2,6) and (3,18) lie on the curve, so you can try substituting them in as the x and y co-ordinates)
3. (Original post by Jarred)
Have you tried forming a system of two simultaneous equations in terms of a and n?

(Hint: The points (2,6) and (3,18) lie on the curve, so you can try substituting them in as the x and y co-ordinates)
I have tried - but not succeeded.
I ended up with something like (nlog18 = log3 - loga) - (nlog6 = log2 - loga) but didnt know how to progress further than that. I get log2 = log(3/2) ... which makes no sense what so ever
4. You have to graph logY again LogX

x 2 3
y 6 18
logx 0.30103 0.4771
logy 0.7782 1.2553

You turn y=ax^n into a leaner form using logs:
y = m*x + c
logy = n*logx + loga

Find the gradient to find n, find the y intercept to find a.
Hope this helps
5. (Original post by KanKan)
I have tried - but not succeeded.
I ended up with something like (nlog18 = log3 - loga) - (nlog6 = log2 - loga) but didnt know how to progress further than that. I get log2 = log(3/2) ... which makes no sense what so ever
How do you get that? Subtracting your 2 equations should give you your value for n.
6. (Original post by KanKan)
I have tried - but not succeeded.
I ended up with something like (nlog18 = log3 - loga) - (nlog6 = log2 - loga) but didnt know how to progress further than that. I get log2 = log(3/2) ... which makes no sense what so ever
Here are the equations I got:

From here, you just need to use the normal method of elimination. If you minus one equation from the other, you get rid of the log(a) term and can rearrange to find n.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 14, 2013
Today on TSR

### What is the latest you've left an assignment

And actually passed?

### Simply having a wonderful Christmas time...

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

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

• Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

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

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.