Algebra 2 help Please find attachment

Half-life of $15$ years means that the amount of DDT in the soil halves every $15$ years.

2. Let $A_0 = 0.5$ mg/kg be the amount of DDT in the soil in 1972. Then the formula for the amount of DDT in the year $1972 + t$ is given by \[ A(t) = A_0 \cdot 2^{-t/15}. \]

1-3. This gives us the following table, formatted as (year, concentration): just look for your birth year to get the value you need.

(1972, 0.5000), (1973, 0.4774), (1974, 0.4559), (1975, 0.4353), (1976, 0.4156), (1977, 0.3969), (1978, 0.3789), (1979, 0.3618), (1980, 0.3455), (1981, 0.3299), (1982, 0.3150), (1983, 0.3008), (1984, 0.2872), (1985, 0.2742), (1986, 0.2618), (1987, 0.2500), (1988, 0.2387), (1989, 0.2279), (1990, 0.2176), (1991, 0.2078), (1992, 0.1984), (1993, 0.1895), (1994, 0.1809), (1995, 0.1727), (1996, 0.1649), (1997, 0.1575), (1998, 0.1504), (1999, 0.1436), (2000, 0.1371), (2001, 0.1309), (2002, 0.1250), (2003, 0.1194), (2004, 0.1140), (2005, 0.1088), (2006, 0.1039), (2007, 0.09921), (2008, 0.09473), (2009, 0.09045), (2010, 0.08637), (2011, 0.08247), (2012, 0.07874), (2013, 0.07519), (2014, 0.07179), (2015, 0.06855), (2016, 0.06546), (2017, 0.06250), (2018, 0.05968), (2019, 0.05698), (2020, 0.05441), (2021, 0.05195), (2022, 0.04961)


4. To find the year when the concentration falls below $0.008$ mg/kg, we have to find $t$ such that \[ 0.5 \cdot 2 ^{-t/15} \leq 0.008\] or equivalently \[ -\frac{t}{15} \leq \log_2\left(\frac{0.008}{0.5}\right) \] or again \[ t \geq -15 \cdot \log_2(0.016) \simeq (-15)(-5.96) = 89.4, \] which means $90$ years after $1972$, so in $2062$.

I think that for part 1. you are only supposed to take the years that are 15-30-45-etc. years after 1972, so 1972 itself, 1987, 2002, 2017. That's because you don't need the formula to compute the concentration, just halve it every time. But not sure about that.