Question 1195759
.
The radioactive element​ carbon-14 has a​ half-life of 5750 years. A scientist determined that the bones 
from a mastodon had lost 60.4​% of their​ carbon-14. How old were the bones at the time they were​ discovered?
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On solving similar problem, related to radioactive decay, see the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/logarithm/Radioactive-decay-problems.lesson>Radioactive decay problems</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/logarithm/Carbon-dating-problems.lesson>Carbon dating problems</A> 

in this site.


Learn the subject from there.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Logarithms</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.



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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The solution by @josgaithmetic in his post is &nbsp;INCORRECT &nbsp;from its first line 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;to the last line including the answer, &nbsp;which is &nbsp;WRONG, &nbsp;too.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In my post, I will bring a correct solution, &nbsp;and then at the end, &nbsp;will explain,
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;why everything in post by @josgarithmetic is wrong.



<pre>
Since you are given half-life period, you do not need to write the exponential
decay function in the (ekt)-form and then to determine the parameter "k" of this form.


Write exponential decay function for the remaining mass using base 1/2 in the form

    M(t) = {{{M(0)*(1/2)^(t/5750)}}}.    (1)


It is the STANDARD exponential decay form function, when the half-life is given.


Next, in your problem, the lost mass of the carbon-14 is 60.4%.

It means that remaining mass of carbon-14 is 100%-60.4% = 39.6%.

THEREFORE, the equation (1) for the remaining mass takes the form

    0.396 = {{{(1/2)^(t/5750)}}},


or, which is the same,

    0.396 = 0.5^(t/5750).    (2)


Take logarithm base 10 of both sides of (2).  You will get

    log(0.396) = {{{(t/5750)*log((0.5))}}}


which implies

    t = {{{5750*(log((0.396))/log((0.5)))}}} = 7684 years, or, rounding to closest one hundred years,

as usually people do in such calculations, we get the 


<U>ANSWER</U>.  The bones are 7700 years old.
</pre>

Solved.


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The solution by @josgarithmetic has two errors.


First error is that he starts from (ekt)-form of the exponential decay function,
having given data about the half life period.


In such cases, &nbsp;the normal procedure is to use the base &nbsp;1/2 &nbsp;exponential
function. &nbsp;Otherwise, &nbsp;you will make unnecessary calculations to find the coefficient &nbsp;"k"
of the exponential function, &nbsp;and, &nbsp;at the same time, &nbsp;you will demonstrate
that you are infamiliar with the regular procedure.


The second error in the solution by @josgarithmetic is that he uses given &nbsp;60.4% 
as the remaining mass, &nbsp;although the problem clearly states that it is the lost mass percentage.



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    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Everything related to the solution procedure for exponential equation,
    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;when the half life is given, is described and clearly explained 
    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;in my lesson/lessons, &nbsp;to which &nbsp;I &nbsp;referred in my post above.


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After my post, @josgarithmetic changed/corrected his calculations, related to the remaining mass.