The age of a rock is determined by the function, A(t) = A 0 e ^(-0.000002t), where Isotope Q, with an initial value of A 0, decays at a constant rate 0.000002 per year, for a number of years (also known as the age of the rock), t. A sample originally has 80 grams of Isotope Q, decays, and now has 20 grams of the isotope. Determine the current age of the sample, in years. Round your answer to the nearest thousand years. Make sure to include the correct place value commas in your answer. The age of the sample ≈ ____ years.

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The age of a rock is determined by the function, A(t) = A 0 e ^(-0.000002t), where Isotope Q, with an initial value of A 0, decays at a constant rate 0.000002 per year, for a number of years (also known as the age of the rock), t. A sample originally has 80 grams of Isotope Q, decays, and now has 20 grams of the isotope. Determine the current age of the sample, in years. Round your answer to the nearest thousand years. Make sure to include the correct place value commas in your answer. The age of the sample ≈ ____ years.

Mathematics
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So: \[A(t)=A_o e ^{0.000002t}\] A function always determines the relationship between two variables which are often labelled "xoy", that is what we call a "reference system", in this case it is "toA" because those are the variables we are dealing with. In any reference system, there is always a dependant variable and a "independant variable", in this case, "t" is the independant variable and "A" is the dependant. But we are given a "dependant variable" value, so we have to find the value of the independant variable, which in this case is the time. But since it's the dependant value, means that it depends on "t" which implies that the value we are given is for sure A(t), so therefore, we only replace the given data: \[20=80e ^{0.000002t}\] And now, we are in front of an univariable equation, for which we have to solve fot "t".
i got t=-693147
May I ee what you did?

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Well, that is a quatumn problem, but I had a typo error, I did not see that the exponent is negative. So, if you simplify it by isolating the e^-0.000002t: \[\frac{ 20 }{ 80 }=e ^{-0.000002t}\] And taking the natural logarithm on both sides: \[(-0.000002t)Lne=Ln \frac{ 20 }{ 80 }\] And since Lne=1 and isolating the "t": \[t=\frac{ Ln(\frac{ 20 }{ 80 }) }{ -0.000002 }\] Thing is, it is absurd for time to be negative, you don't really need to know details about that, but there's no way that isotope can go back in time, so we want the absolute value of this expression: \[t=\left| \frac{ Ln (\frac{ 20 }{ 80 }) }{ -0.000002t } \right|\]

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