Turn on thread page Beta
    • Thread Starter
    Offline

    12
    ReputationRep:
    Hello I am really struggling with this question, wonder could anyone solve this for me?
    log6 + log(x-3) =2 log y
    2y -x =3

    thanks


    ===Moved to Maths===
    Offline

    13
    ReputationRep:
    (Original post by Marcus2016)
    Hello I am really struggling with this question, wonder could anyone solve this for me?
    log6 + log(x-3) =2 log y
    2y -x =3

    thanks


    ===Moved to Maths===
    I'm not going to solve it for you, but I will give you a starting step/hint:

    Can you use the equation  \displaystyle \log 6 + \log \left(x-3\right) = 2 \log y to express  y in terms of  x, or just an equation linking the two?

    Then it's just solving simultaneous equations from C1!!
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by crashMATHS)
    I'm not going to solve it for you, but I will give you a starting step/hint:

    Can you use the equation  \displaystyle \log 6 + \log \left(x-3\right) = 2 \log y to express  y in terms of  x, or just an equation linking the two?

    Then it's just solving simultaneous equations from C1!!
    Thanks but when i multiply log into the bracket what does it become, is it log -3, as you can't get a negative log?
    Offline

    20
    ReputationRep:
    remember....

    log p + log q = log pq

    alog p = log {pa}
    Offline

    13
    ReputationRep:
    (Original post by Marcus2016)
    Thanks but when i multiply log into the bracket what does it become, is it log -3, as you can't get a negative log?
    Oh no, you have a misunderstanding. It is generally not true that  \log (a+b) = \log(a) + \log(b) . That is not how logarithms work.

    What you want to be using is that:  \log(a) + \log(b) = \log(ab) for the left hand side. In this case, let your  a = 6 , b = x-3
    • Thread Starter
    Offline

    12
    ReputationRep:
    Thanks everyone question solved I owe you one!
    • Thread Starter
    Offline

    12
    ReputationRep:
    sorry to ask again but any hints for the following?

    2 + log2(2x +1) =2 log2 y
    x=22-y
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Marcus2016)
    sorry to ask again but any hints for the following?

    2 + log2(2x +1) =2 log2 y
    x=22-y
    2=2\log_2(2)

    Then apply log rules and get rid off the logs.
    Offline

    13
    ReputationRep:
    (Original post by Marcus2016)
    sorry to ask again but any hints for the following?

    2 + log2(2x +1) =2 log2 y
    x=22-y
    RDKgames has suggested a nice method or you may like to rearrange the first equation to:

     \displaystyle 2 \log_2 \left(y\right) - \log_2 \left(2x+1\right) = 2 and use the rule that  \displaystyle \log(a) - \log(b) = \log\left(\frac{a}{b}\right)
    • Thread Starter
    Offline

    12
    ReputationRep:
    Thanks again!!!
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 13, 2017
Poll
Favourite type of bread
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

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

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