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Log Functions

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  • 14-04-2015 1:11pm
    #1
    iCrazzy
    Registered Users Posts: 204 ✭✭


    Could someone explain to me how to do these 2 qs.
    8 and 12


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    #2
    RoRo979
    Registered Users Posts: 476 ✭✭RoRo979


    hmm well q8 not a clue but i may have an idea for q12, basically will end up with a simultaneous equations. rearrange the line and youll get

    y-b= alog2x ( divide by a)
    y-b / a = log2x ( sub in numbers)
    10 - b/a = log2^8 ( which is 3)

    10 -b/a = 3 ( multiplu by a)

    10 - b = 3a

    3a + b = 10

    do same for other points and solve


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    #3
    A97
    Registered Users Posts: 390 ✭✭A97


    Both end up as simultaneous equations. For the first one, plug in the corresponding x and y values, (0,0) and (1,14). Sub in the two x and two y values in the second one too, (8,10) and (32,14).

    See the attached solutions. (Doing maths on paint is not fun. :P)

    Now for some Gaeilge. :)


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    #4
    iCrazzy
    Registered Users Posts: 204 ✭✭iCrazzy


    A97 wrote: »
    Both end up as simultaneous equations. For the first one, plug in the corresponding x and y values, (0,0) and (1,14). Sub in the two x and two y values in the second one too, (8,10) and (32,14).

    See the attached solutions. (Doing maths on paint is not fun. :P)

    Now for some Gaeilge. :)

    Hi, thanks, I understand Q 12, but not how you did Q7.


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    #5
    Peregrine
    Moderators, Education Moderators Posts: 26,399 Mod ✭✭✭✭Peregrine


    iCrazzy wrote: »
    Hi, thanks, I understand Q 12, but not how you did Q7.

    Is it Q8 or Q7 that you wanted?

    Q8 and Q12 are more or less the same question. Q12 explicitly gives you two points that satisfy the equation where as Q8 gives you a graph of the function.

    From the graph, we know that the point (1,14) satisfies the equation. Even thought it's not stated, we also know that it goes through the origin (0,0). These two points satisfy the equation in Q8.

    We input the x and y values of these two points to create two simultaneous equations with two unknown variables. Remember, e is a known value (2.718..).

    y = ae^x + b
    (0,0)
    0 = ae^0 + b
    Any number to the power of zero is equal to 1.
    a + b = 0

    (1,14)
    14 = ae^1 + b
    ea + b = 14

    I don't know how you usually solve simultaneous equations but you should be able to solve them from there.


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    #6
    iCrazzy
    Registered Users Posts: 204 ✭✭iCrazzy


    Nim wrote: »
    Is it Q8 or Q7 that you wanted?

    Q8 and Q12 are more or less the same question. Q12 explicitly gives you two points that satisfy the equation where as Q8 gives you a graph of the function.

    From the graph, we know that the point (1,14) satisfies the equation. Even thought it's not stated, we also know that it goes through the origin (0,0). These two points satisfy the equation in Q8.

    We input the x and y values of these two points to create two simultaneous equations with two unknown variables. Remember, e is a known value (2.718..).

    y = ae^x + b
    (0,0)
    0 = ae^0 + b
    Any number to the power of zero is equal to 1.
    a + b = 0

    (1,14)
    14 = ae^1 + b
    ea + b = 14

    I don't know how you usually solve simultaneous equations but you should be able to solve them from there.


    y = ae^x + b
    (0,0)
    0 = ae^0 + b
    Any number to the power of zero is equal to 1.
    a + b = 0

    (1,14)
    14 = ae^1 + b
    ea + b = 14

    ea+b=14

    I don't think you can do that way


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    #7
    Peregrine
    Moderators, Education Moderators Posts: 26,399 Mod ✭✭✭✭Peregrine


    iCrazzy wrote: »
    I don't think you can do that way

    Why not?

    From there, you can solve for a and b using substitution or elimination, whichever you prefer.


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    #8
    iCrazzy
    Registered Users Posts: 204 ✭✭iCrazzy


    Nim wrote: »
    Why not?

    From there, you can solve for a and b using substitution or elimination, whichever you prefer.

    a + b = 0
    ea+b=14

    How can you do simulatenous equations in that ?


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    #9
    Peregrine
    Moderators, Education Moderators Posts: 26,399 Mod ✭✭✭✭Peregrine


    iCrazzy wrote: »
    a + b = 0
    ea+b=14

    How can you do simulatenous equations in that ?

    Multiply the first equation by -1.

    [latex]-a -b = 0[/latex]
    [latex]ea+b=14[/latex]

    Eliminate the [latex]b[/latex]s

    [latex]ea - a = 14[/latex]
    [latex]a(e-1) = 14[/latex]
    [latex]a =\frac{14}{e-1}[/latex]

    [latex]a+b = 0[/latex]
    [latex]b = -a[/latex]
    [latex]b = -\frac{14}{e-1} [/latex]

    Or substitute
    [latex]a = -b[/latex] into [latex]ea+b=14[/latex] like A97 did.


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    #10
    iCrazzy
    Registered Users Posts: 204 ✭✭iCrazzy


    Nim wrote: »
    Multiply the first equation by -1.

    [latex]-a -b = 0[/latex]
    [latex]ea+b=14[/latex]

    Eliminate the [latex]b[/latex]s

    [latex]ea - a = 14[/latex]
    [latex]a(e-1) = 14[/latex]
    [latex]a =\frac{14}{e-1}[/latex]

    [latex]a+b = 0[/latex]
    [latex]b = -a[/latex]
    [latex]b = -\frac{14}{e-1} [/latex]

    Or substitute
    [latex]a = -b[/latex] into [latex]ea+b=14[/latex] like A97 did.

    Feel really embarrassed now, my bad.
    I just got confused because ea were joined together and I always thought that you have to have exact same variables to do simultaneous eq.
    But thank you for taking your time to teaching me that is not the case, much appreciated.


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    #11
    Peregrine
    Moderators, Education Moderators Posts: 26,399 Mod ✭✭✭✭Peregrine


    iCrazzy wrote: »
    Feel really embarrassed now, my bad.
    I just got confused because ea were joined together and I always thought that you have to have exact same variables to do simultaneous eq.
    But thank you for taking your time to teaching me that is not the case, much appreciated.

    They are the same variables, just different co-efficients. ea + b = 14 is no different to 2a + b = 14. e is a mathematical constant, like pi, just another number.

    You're welcome :)


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