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c1 question: find the value of y such that 4^y+1=8^2y-1? watch

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    It's a question in one of the solomon papers. I have the answer but I don't understand how it comes to y=5/4?
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    (Original post by funkybinoculars)
    It's a question in one of the solomon papers. I have the answer but I don't understand how it comes to y=5/4?
    Change it, so they're in the same base.
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    (Original post by funkybinoculars)
    It's a question in one of the solomon papers. I have the answer but I don't understand how it comes to y=5/4?
    log everything

    log(4)^y+1 + log(1) = log(8)^2y-1

    use log rules

    y+1log(4) + log(1) = 2y-1log(8)
    log 1 is zero so:
    y+1log(4) = 2y-1log(8)
    divide both sides by log(4) and 2y-1

    y+1/2y-1 = log(8)/log(4)
    y+1/2y-1 = 3/2

    multiply both sides by 2y-1 and 2
    2y+2 = 6y-3
    4y=5
    y=5/4
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    (Original post by TheALevelStudent)
    log everything

    log(4)^y+1 + log(1) = log(8)^2y-1

    use log rules

    y+1log(4) + log(1) = 2y-1log(8)
    log 1 is zero so:
    y+1log(4) = 2y-1log(8)
    divide both sides by log(4) and 2y-1

    y+1/2y-1 = log(8)/log(4)
    y+1/2y-1 = 3/2

    multiply both sides by 2y-1 and 2
    2y+2 = 6y-3
    4y=5
    y=5/4
    yeah i thought about using log but don't know if i'd get a mark because we don't learn logs in c1?
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    (Original post by TheALevelStudent)
    log everything

    log(4)^y+1 + log(1) = log(8)^2y-1

    use log rules

    y+1log(4) + log(1) = 2y-1log(8)
    log 1 is zero so:
    y+1log(4) = 2y-1log(8)
    divide both sides by log(4) and 2y-1

    y+1/2y-1 = log(8)/log(4)
    y+1/2y-1 = 3/2

    multiply both sides by 2y-1 and 2
    2y+2 = 6y-3
    4y=5
    y=5/4
    No need. It is C1. Simple indices will resolve the issue here.

    (Original post by funkybinoculars)
    It's a question in one of the solomon papers. I have the answer but I don't understand how it comes to y=5/4?
    Consider:

     2^x = 2^3

    It shouldn't be difficult to see that x must be 3, because otherwise the sides won't be equal.

    In the case of

     4^{y+1}= 8^{2y-1}

    Consider what base 4 and 8 have in common and convert both of them to that base, i.e if I had 9 I could write it as 3^2.

    Once you have the same base, you know that you powers have to be the same (otherwise you won't have equality).

    Can you go from here?
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    (Original post by funkybinoculars)
    yeah i thought about using log but don't know if i'd get a mark because we don't learn logs in c1?
    (Original post by kingaaran)
    no need. It is c1. Simple indices will resolve the issue here.
    log everything
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    (Original post by TheALevelStudent)
    log everything
    Better get memorising log tables for your C1 exam then
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    (Original post by kingaaran)
    Better get memorising log tables for your C1 exam then
    Lmao, I'm doing A2, so there is no need. In A2 we log everything 😂😂
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    (Original post by TheALevelStudent)
    Lmao, I'm doing A2, so there is no need. In A2 we log everything 😂😂
    Log (everything) = life
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    (Original post by kingaaran)
    Log (everything) = life
    You know it haha 😂😂
 
 
 

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