After 548 hours decreases the activity of a radioactive substance to 1/9 of the initial value. What is the half-life of the substance?

Result

t1/2 =  172.9 hours

#### Solution:

$\left(\dfrac{1}{2}\right)^{\dfrac{ 548}{t_{1/2}}} = \dfrac{1}{ 9} \ \\ \ \\ t_{1/2} = 548 \dfrac{ \ln (0.5)}{ \ln(1/9)} = 172.9 \ \text{hours}$

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