# If you could help me with these two questions I would really appreciate it, Thank you! The questions are in the attached file.

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The half -life of a substance is the time it takes for this substance to decay until only half of the original amount is left. For the sample of 100 mg, it would be the time it would take until half of that amount, or 50 mg, is left.

If the amount after t years is given by `A = 100e^(-0.05t)` , then half-life is determined by plugging in A = 50 and solving for t:

`50 = 100e^(-0.05t)`

Divide both sides of the equation by 100:

`1/2 = e^(-0.05t)`

Take natural logarithm (ln) of both sides. This will remove exponent e from the right side:

`ln(1/2) = -0.05t`

According to the rules of the logarithms, `ln(1/2) = -ln2`

-ln2 = -0.05t

From here, `t = (ln2)/0.05 =13.86 ` . Assuming that the constant 0.05 is given in 1/year, the time is 13.86, or about 14 years.

**The half-life of a given substance is 14 years, rounded to the nearest year.**

To find the number of days 100 people are infected by the virus, x, plug in y = 100:

`100 = 1000/(1+999e^(-.3x))`

Multiply both sides of the equation by the denominator:

`100(1+999e^(-.3x)) = 1000`

Divide both sides by 100:

`1+999e^(-.3x) = 10`

Subtract 1 from both sides:

`999e^(-.3x) = 9`

Divide by 999:

`e^(-.3x) = 1/111`

Take natural logarithm of both sides to remove the exponent:

`-.3x = ln(1/111)`

Finally, divide both sides by -.3 to obtain x:

`x = (ln(1/111))/(-.3)`

x = 15.698 or 15.7, if rounded to the nearest tenth.

**It will be 15.7 days, rounded to the nearest tenth, until 100 people are infected with the virus.**

for the 1st question:

A=100^e^-0.05t

so after 0 years which is what youre starting out with t=0

and A=100^e^0

anything raised to to the power 0 is 1 so 100 mg which they said is what you started with

half life is when A=50mg

so 50mg=100^e^-0.05t

0.5=e^-0.05t

ln 0.5=ln (e^-0.05t)

ln0.5=-0.05t lne

ln e =1

ln 0.5 divide by -0.05=t

and you get your answer!