Half-life of Radioactive Substance: Solve Algebraically w/Logarithms

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To determine the half-life of a radioactive substance that decays from 3150g to 450g in 73 weeks, the correct equation to use is 450 = 3150(1/2)^(73/T), where T represents the half-life. Rearranging this equation allows for the calculation of T using logarithms. The solution indicates that the half-life is approximately 26 weeks. The initial approach using 450 = 3150(x)^73 was incorrect. Accurate application of logarithmic functions is essential for solving half-life problems effectively.
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If 3150g of a radioactive substance decays to 450g in 73 weeks, determine the half life of the substance to the nearest week. Solve algebraically using logarithms.

450=3150(x)^73
450/3150=x^73
73sqrt(450/3150)=x

Now, the only problem is the answer says it is 26 weeks. Any help is much appreciated.
 
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You're starting with the wrong equation. Try something like

450 = 3150 \left( \frac {1}{2} \right)^{73/T}

where T is the half-life.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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