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Urgent help please ! Logs / intersections point Watch

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    Need some help please I would be very appreciative of any help. Thanks in advance.


    The curves y = 3^(x+2) and y = 5^(x-1) meet at a point Find the point of intersection of where the two curves meet.

    Edit: I'm sooo sorry I made a typo, it should have been this instead 😁😩
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    (Original post by Aty100)
    Need some help please I would be very appreciative of any help. Thanks in advance.


    The curves y = 3^(x+1) and y = 5^(x-1) meet at a point Find the point of intersection of where the two curves meet.
    solve their equations simulatenously
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    (Original post by TeeEm)
    solve their equations simulatenously
    With logs?

    What is did was equated them and then did log base of 3 and solved
    I got X to be -2.738
    I don't think that works :/
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    (Original post by Aty100)
    With logs?

    What is did was equated them and then did log base of 3 and solved
    I got X to be -2.738
    I don't think that works :/
    That's not right. Can you post all your working?
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    (Original post by notnek)
    That's not right. Can you post all your working?
    Thanks for helping
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    (Original post by Aty100)
    Need some help please I would be very appreciative of any help. Thanks in advance.


    The curves y = 3^(x+1) and y = 5^(x-1) meet at a point Find the point of intersection of where the two curves meet.
     (x+1) \log_3{3} = (x-1) \log_3{5}

     x(1 - \log_3{5}) = -1 - \log_3{5}

     x = - \dfrac{1 + \log_3{5}}{1 - \log_3{5}}

    Remember that you can still use the log rules even if the bases are weird like this one. Maybe it would be neater to have done it with natural logs.
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    (Original post by Aty100)
    Thanks for helping
    I'm afraid that is not the question that you posted originally.
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    (Original post by Aty100)
    Thanks for helping
    That's a bit different to the question you posted - which one is correct?

    There's a mistake in your working:

    x + 2 - x\log_3 5 = x(2-log_3 5)

    If you expand the right-hand-side you'll see that this is wrong.

    Instead, move the 2 to the other side and then take out a factor of x.
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    (Original post by Louisb19)
     (x+1) \log_3{3} = (x-1) \log_3{5}

     x(1 - \log_3{5}) = -1 - \log_3{5}

     x = - \dfrac{1 + \log_3{5}}{1 - \log_3{5}}

    Remember that you can still use the log rules even if the bases are weird like this one. Maybe it would be neater to have done it with natural logs.
    OMG I TYPED MY QUESTION OUT WRONG!!
    I'm Sooo sorry.

    It's supposed to be 3^(X+2)
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    (Original post by notnek)
    That's a bit different to the question you posted - which one is correct?

    There's a mistake in your working:

    x + 2 - x\log_3 5 = x(2-log_3 5)

    If you expand the right-hand-side you'll see that this is wrong.

    Instead, move the 2 to the other side and then take out a factor of x.
    Sorry it should have been 3^x+2
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    (Original post by Aty100)
    Sorry it should have been 3^x+2
    Okay. My last post explains where you went wrong.
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    (Original post by notnek)
    Okay. My last post explains where you went wrong.

    I don't understand why it is x(2-log3 5)
    Where did the 2 come from ?
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    (Original post by Aty100)
    I don't understand why it is x(2-log3 5)
    Where did the 2 come from ?
    You wrote x(2-log3 5) in your working.

    I'm telling you that it's incorrect.
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    (Original post by notnek)
    You wrote x(2-log3 5) in your working.

    I'm telling you that it's incorrect.
    Oh sorry.

    So

    X+2 = x(log3 5) - 1 ( log3 5)

    X - Xlog3 5 = -1log3 5 -2

    X(1-log3 5) = ( -1log3 5 ) - 2

    X = ( -1log3 5 ) - 2 / (1-log3 5)

    ??
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    (Original post by Aty100)
    Need some help please I would be very appreciative of any help. Thanks in advance.


    The curves y = 3^(x+2) and y = 5^(x-1) meet at a point Find the point of intersection of where the two curves meet.

    Edit: I'm sooo sorry I made a typo, it should have been this instead 😁😩
    As suggested above it would be easier to take logs base 10, rather than stressing and getting confused about log base 3
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    (Original post by Aty100)
    Oh sorry.

    So

    X+2 = x(log3 5) - 1 ( log3 5)

    X - Xlog3 5 = -1log3 5 -2

    X(1-log3 5) = ( -1log3 5 ) - 2

    X = ( -1log3 5 ) - 2 / (1-log3 5)

    ??
    That's correct.
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    (Original post by notnek)
    That's correct.
    I got X= 7.4 but that doesn't work when I sub it back into the y equations 😩
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    (Original post by Aty100)
    Oh sorry.

    So

    X+2 = x(log3 5) - 1 ( log3 5)

    X - Xlog3 5 = -1log3 5 -2

    X(1-log3 5) = ( -1log3 5 ) - 2

    X = ( -1log3 5 ) - 2 / (1-log3 5)

    ??

    For an alternative method
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    (Original post by Aty100)
    I got X= 7.4 but that doesn't work when I sub it back into the y equations 😩
    It's correct and it satisfies the equations if you use the exact value of x i.e. not the rounded value 7.4. (actually 7.5 is correct for 1 d.p.)
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    (Original post by notnek)
    It's correct and it satisfies the equations if you use the exact value of x i.e. not the rounded value 7.4. (actually 7.5 is correct for 1 d.p.)
    Thank SOO much !
 
 
 
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