Question 1201501
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Answer: <font color=red size=4>102.69 years</font>


Work Shown:


x = number of years
y = amount of substance leftover, in grams


One possible equation is
y = 6000*(0.5)^(x/18)
it is of the format
y = a*(0.5)^(x/H)
where 'a' is the starting amount and H is the half-life in years.


Plug in y = 115 and solve for x.
We'll need to use logs to isolate the exponent.
If the variable is in the trees, then we log it down.


y = 6000*(0.5)^(x/18)
115 = 6000*(0.5)^(x/18)
115/6000 = (0.5)^(x/18)
0.0191667 = (0.5)^(x/18)
Log(0.0191667) = Log( (0.5)^(x/18) )
Log(0.0191667) = (x/18)*Log( 0.5 ) ... use the <a href="https://www.rapidtables.com/math/algebra/Logarithm.html">logarithm power rule</a>
x/18 = Log(0.0191667)/Log( 0.5 )
x = 18*Log(0.0191667)/Log( 0.5 )
x = 102.694576057311
x = <font color=red>102.69</font>
It takes roughly <font color=red>102.69 years</font> to have 115 grams of substance remaining. 



Another problem involving half-life.
<a href = "https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1201494.html">https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1201494.html</a>
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