Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    Please help me with this, I've tried for way too long on this, dont ask me to show my working because I've already wasted 1 hour on this and don't want to waste another minute now. Thanks.
    Attached Images
     
    Offline

    7
    ReputationRep:
    ignore me im an idiot

    Posted from TSR Mobile
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by Jas1947)
    Please help me with this, I've tried for way too long on this, dont ask me to show my working because I've already wasted 1 hour on this and don't want to waste another minute now. Thanks.
    If you let y=2^x, the equation becomes y^2-8y+15=0 which you can solve.

    Once you have y=... it will be the equivalent to 2x=... which you can just use logs to find what what the two values of x are. Add those two roots together and you're done.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    If you let y=2^x, the equation becomes y^2-8y+15=0 which you can solve.

    Once you have y=... it will be the equivalent to 2x=... which you can just use logs to find what what the two values of x are. Add those two roots together and you're done.
    Oh my god i remember now! I cant believe i wasted so much time, thank you!
    Offline

    7
    ReputationRep:
    (Original post by RDKGames)
    If you let y=2^x, the equation becomes y^2-8y+15=0 which you can solve.

    Once you have y=... it will be the equivalent to 2x=... which you can just use logs to find what what the two values of x are. Add those two roots together and you're done.
    I was trying to figure out where I went wrong, I thought product meant multiplication so I was multiplying the roots xD

    Edit: read the question again and it says sum and not product. ok then im seeing things

    Posted from TSR Mobile
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by HFancy1997)
    I was trying to figure out where I went wrong, I thought product meant multiplication so I was multiplying the roots xD

    Edit: read the question again and it says sum and not product. ok then im seeing things

    Posted from TSR Mobile
    Take a break from maths at 1am, mate.
    Offline

    22
    ReputationRep:
    For a more elegant method, y = 2^x gives y^2 - 8y + 15 = 0 so that 2^{\alpha} \cdot 2^{\beta} = 15 where \alpha, \beta are the roots. Then taking \log_2 gives \alpha + \beta = \log_2 15
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by Zacken)
    For a more elegant method, y = 2^x gives y^2 - 8y + 15 = 0 so that 2^{\alpha} \cdot 2^{\beta} = 15 where \alpha, \beta are the roots. Then taking \log_2 gives \alpha + \beta = \log_2 15
    So that's how you do it. Back then at midnight I started thinking how the heck I could apply the alpha beta stuff to this and you've shown it, so thanks.
    Offline

    22
    ReputationRep:
    (Original post by RDKGames)
    So that's how you do it. Back then at midnight I started thinking how the heck I could apply the alpha beta stuff to this and you've shown it, so thanks.
    No problem. Motivating thing here is that product of roots of a polynomial is easy and logarithms turn products into sums.
 
 
 
  • 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
    What's your favourite Christmas sweets?
    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
  • 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

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