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Math Help - Logarithm help!

  1. #1
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    Logarithm help!

    Hi all, I'm having a hard time figuring out how to do the last part of my homework, I would be grateful if someone could explain to me how to do this.

    A cup of tea is cooling in the shade. T degrees Celsius, a spreadsheet was used to plot log e ( T ) against time, t, in minutes.
    The spreadsheet calculated the line of best fit to have an equation expressed as Log e (T) = -0.045t + 4.4543

    e is ~2.71 on the calculator.

    1) Show that in exponential form this is equivalently written as T = e ^ (-0.045t + 4.4543).
    I'm aware that using the Log law you can write that out in 1 step, but not sure how to prove it. Because going from
    Log e (T) = -0.045t + 4.4543 ----> T = e ^ (-0.045t + 4.4543)
    is done in 1 step.

    2) Hence write the expression in the form T=Ae ^ -0.045t
    Show that A=86 to the nearest correct integer.
    Do i just substitute 86 in to A ?

    3)The rule for T can also be written in the form T = 86 * 2 ^ -kt
    a) If T = 86 * 2 ^ -kt =
    86e ^ -0.045t
    Find the value of K to 3 decimals
    I actually have NO IDEA how to do this one ?!

    b) Hence write the time taken for the Excess temperature (The initial temperature which is 86) to be halved and show that
    K = 1/H
    Let H be the Time.

    If anyone could help me out that would be great, I'm going to need to learn this for my up-coming test in a week. Thanks everyone.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Xcel View Post
    Hi all, I'm having a hard time figuring out how to do the last part of my homework, I would be grateful if someone could explain to me how to do this.

    A cup of tea is cooling in the shade. T degrees Celsius, a spreadsheet was used to plot log e ( T ) against time, t, in minutes.
    The spreadsheet calculated the line of best fit to have an equation expressed as Log e (T) = -0.045t + 4.4543

    e is ~2.71 on the calculator.

    1) Show that in exponential form this is equivalently written as T = e ^ (-0.045t + 4.4543).
    I'm aware that using the Log law you can write that out in 1 step, but not sure how to prove it. Because going from
    Log e (T) = -0.045t + 4.4543 ----> T = e ^ (-0.045t + 4.4543)
    is done in 1 step.
    This is because it is essentialy the definition of log to the base e that:

    if y=\log_e(x), then by definition of the \log_e function e^y=x

    2) Hence write the expression in the form T=Ae ^ -0.045t
    Show that A=86 to the nearest correct integer.
    Do i just substitute 86 in to A ?
    You have:

    T = e ^ {-0.045t + 4.4543}

    and by the law of exponents this means:

    T =e^{4.4543} ~e ^ {-0.045t}=85.996 ~e ^ {-0.045t} \approx 86~e ^ {-0.045t}

    3)The rule for T can also be written in the form T = 86 * 2 ^ -kt
    a) If T = 86 * 2 ^ -kt =
    86e ^ -0.045t
    Find the value of K to 3 decimals
    I actually have NO IDEA how to do this one ?!

    b) Hence write the time taken for the Excess temperature (The initial temperature which is 86) to be halved and show that
    K = 1/H
    Let H be the Time.

    If anyone could help me out that would be great, I'm going to need to learn this for my up-coming test in a week. Thanks everyone.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Xcel View Post
    Hi all, I'm having a hard time figuring out how to do the last part of my homework, I would be grateful if someone could explain to me how to do this.

    A cup of tea is cooling in the shade. T degrees Celsius, a spreadsheet was used to plot log e ( T ) against time, t, in minutes.
    The spreadsheet calculated the line of best fit to have an equation expressed as Log e (T) = -0.045t + 4.4543

    e is ~2.71 on the calculator.

    1) Show that in exponential form this is equivalently written as T = e ^ (-0.045t + 4.4543).
    I'm aware that using the Log law you can write that out in 1 step, but not sure how to prove it. Because going from
    Log e (T) = -0.045t + 4.4543 ----> T = e ^ (-0.045t + 4.4543)
    is done in 1 step.

    2) Hence write the expression in the form T=Ae ^ -0.045t
    Show that A=86 to the nearest correct integer.
    Do i just substitute 86 in to A ?

    3)The rule for T can also be written in the form T = 86 * 2 ^ -kt
    a) If T = 86 * 2 ^ -kt =
    86e ^ -0.045t
    Find the value of K to 3 decimals
    I actually have NO IDEA how to do this one ?!
    By definition of the \log_e function 2=e^{\log_e(2)} so:

    <br />
T = 86 \times 2 ^ {-kt} = 86 (e^{\log_e(2)})^{-kt}=86~e^{-k~\log_e(2)~t}<br />

    so if this is the same as the earlirt cooling law:

    <br />
k~\log_e(2)=0.045<br />

    or:

    <br />
k=0.045/\log_e(2)<br />

    b) Hence write the time taken for the Excess temperature (The initial temperature which is 86) to be halved and show that
    K = 1/H
    Let H be the Time.

    If anyone could help me out that would be great, I'm going to need to learn this for my up-coming test in a week. Thanks everyone.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Xcel View Post
    Hi all, I'm having a hard time figuring out how to do the last part of my homework, I would be grateful if someone could explain to me how to do this.

    A cup of tea is cooling in the shade. T degrees Celsius, a spreadsheet was used to plot log e ( T ) against time, t, in minutes.
    The spreadsheet calculated the line of best fit to have an equation expressed as Log e (T) = -0.045t + 4.4543

    e is ~2.71 on the calculator.

    1) Show that in exponential form this is equivalently written as T = e ^ (-0.045t + 4.4543).
    I'm aware that using the Log law you can write that out in 1 step, but not sure how to prove it. Because going from
    Log e (T) = -0.045t + 4.4543 ----> T = e ^ (-0.045t + 4.4543)
    is done in 1 step.

    2) Hence write the expression in the form T=Ae ^ -0.045t
    Show that A=86 to the nearest correct integer.
    Do i just substitute 86 in to A ?

    3)The rule for T can also be written in the form T = 86 * 2 ^ -kt
    a) If T = 86 * 2 ^ -kt =
    86e ^ -0.045t
    Find the value of K to 3 decimals
    I actually have NO IDEA how to do this one ?!

    b) Hence write the time taken for the Excess temperature (The initial temperature which is 86) to be halved and show that
    K = 1/H
    Let H be the Time.
    You now have:

    <br />
T= 86 \times 2^{-kt}<br />

    When t=1/k this becomes:

    <br />
T(t=1/k) = 86 \times 2^{-1}= \frac{86}{2}<br />

    and as 86 is the temprature at t=0 the temprature has halved when t=1/k

    RonL
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  5. #5
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    Much appreciated <3 Thank you!
    One final question. How can I make LOG to the base "e" on my grapics calculator. Since it's automatically LOG 10, how to change to LOG e?
    Last edited by Xcel; September 5th 2007 at 11:42 PM.
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  6. #6
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    Quote Originally Posted by Xcel View Post
    Much appreciated <3 Thank you!
    One final question. How can I make LOG to the base "e" on my grapics calculator. Since it's automatically LOG 10, how to change to LOG e?
    Your calculator should have a button "ln", this is log base e. (The name for it is "natural logarithm.")

    -Dan
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