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How do i find the constant in a half life equation?

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Hello,
I have no idea what can of data you have, so I am taking a theoretical stance. In a radioactive decay the period of half life is the time after which half the nuclei have desintegrated. Let us call that period tau τ.
The constant, usually called lambda, λ appears in the expoential. It is the constant term that multiplies the time variable, t.
If you have y1 nuclei at a time t, and y2 nuclei at time t+ τ., the ratio y2/y1 =1/2. It is also equal to e^(-λ* τ). Thus
e^(-λ* τ) =1/2.
and ln(e^(-λ* τ)) = ln(1/2)=-ln(2). But since ln(e¨(x)) = x, we get
-λ* τ= - ln(2) and
λ= ln(2)/ τ

In the enclosed screen capture the equation e^(-λ* τ) =1/2 is solved for λ
how do i find the constant in a half life - 12470f3.jpg
If you have experimental data, or a radioactive decay curve drawn you can obtain the constant λ , as the negative of the slope of the function at t=0. See on following graph. Disregard the values of the function before t=0 ( I should have restricted the window to the first quadrant.) The function drawn is e^(-0.365t)

4103394.jpg
Hope it helps.
Thank you for using FixYa

Posted on Nov 20, 2009

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a6ea72a6-0c88-4d79-965a-e93ed8980d8f.png

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Definition



A mathematical statement used to evaluate a value. An equation can use any combination of mathematical operations, including addition, subtraction, division, or multiplication. An equation can be already established due to the properties of numbers (2 + 2 = 4), or can be filled solely with variables which can be replaced with numerical values to get a resulting value. For example, the equation to calculate return on sales is: Net income ÷ Sales revenue = Return on Sales. When the values for net income and sales revenue are plugged into the equation, you are able to calculate the value of return on sales.

There are many types of mathematical equations.

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I can help you there :) In the upper-left hand corner of your keyboard, below the touch pad, there should be a button labelled '=' Enter the first part of your equation press '=' then enter the second half. After you have entered the equation hit enter and it will tell you if it is true or false.

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For this type of problem, a equals the constant cost of operation (monthly rent..etc) and b equals the cost to manufacture the items.

first equation: 320 = a + b(10)
second equation: 520 = a +b(20)

320 = a + 10b
520 = a + 20b subtract equation 2 from equation 1

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200 = -10b divide by -10

20 = b plug b into one of the equations

320 = a +(20)(10) multiply

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How do I use the solve function


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PART 1 a_n*X+b_n*Y=C_n.

In this type, you have only 2 unknows, X, Y, the others are coefficients. When this mode is selected, you are given access to a what looks like a matrix.
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The first column, also inaccessible to you numbers the lines as (1 and 2)
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Enter the coefficient of Y, (b_1) in the first equation and press [=] Cursor moves to the dark rectangle under heading c.
Enter the constant term in the first equation, C_1, and press [=]

At this stage you have finished entering the coefficients and constant term of the first linear equation of your system, namely
a_1*X+b_1*Y= C_1

After you enter C_1 and press [ENTER] the cursor moves to the

Second Line

On this line, just as in the previous one, you enter a_2, b_2 and c_2 Once you finish entering all coefficients and constant terms you press [ENTER].

Calculator does its thing and returns with a solution and displays at the top left of the screen
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Press [=] again and the screen displays the value of the second unknown Y.
If you press [=] a third time the claculator takes you back to the entry screen to modifiy coefficients and constant terms.

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When finished you have to press [=] 1 time to get X, a second time to get Y, and a third time to get Z.

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PS to the power that be of this site, if you are listening: We ought to have prompts to confirm the post before validating it. We ought to be able to correct typos, errors; to amend what we entered.

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1 Answer

A(24)=40,000e((ln2/26)24)


your best bet in any kind of exponential equation is to set up the general form of the equation and then solve for the constants with the numbers they gave you. With P being population, t being time, you know it is going to look like P=Ae^(b*t) and you are given 2 P's for 2 t's.

So for the first set :

Eq 1 P(26)=40,000=Ae^(b*26)

and the second being a time t=0 P was half the current value:

Eq 2 P(0)=20,000=Ae^(b*0)
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now b from Eq 1

40,000=20,000e^(b*26)






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