Lauren B. answered • 10/07/19

Enthusiastic Physics, Chemistry and Math Tutor with 10yrs experience

So if it is reduced by 40% in 35 hours then it has 60% remaining after 35 hours. Using our basic formula

A=A_{0}e^{kt}

If our starting amount (A_{0}) is 100 then our amount remaining (A) is 60, if our starting amount (A_{0}) is 1 then our amount remaining (A) is 0.6. We can also just leave A_{0} and use A=0.6(A_{0}). Whatever makes the most sense to you.

Using A=0.6(A_{0}), t=35

0.6(A_{0})=A_{0}e^{k(35)}

Divide both sides by A_{0 }(This step whatever choice you made for your starting amount will cancel out and we will get the same result.

0.6=e^{k(35)}

Take the ln of both sides

ln(0.6)=35k

divide by 35

k=ln(0.6)/35

plug this into equation

A=A_{0}e^{(ln(0.6)/35)t}

Now that we have the expression we can solve for t when A=0.5A_{0} (Or A=0.5 A_{0}=1 Or A=50 A_{0}=100)

0.5A_{0}=A_{0}e^{(ln(0.6)/35)t}

Follow same steps as above but this time we are solving for t

0.5=e^{(ln(0.6)/35)t}

Take ln of each

ln(0.5)=(ln(0.6)/35)t

Multiplty by 35, divide by ln(0.6)

35ln(0.5)/ln(0.6)=t

t≈47.5hrs

You could have found k and rounded to a decimal and used that instead of tracking through ln(0.6)/35 if you prefer.

Rebecca M.

*Would the answer for k be -0.0146?10/07/19

Lauren B.

10/07/19

Rebecca M.

so to find my answer would I just plug in that number to my first equation?10/07/19

Rebecca M.

and where are you getting the 0.5 from?10/07/19

Lauren B.

10/07/19

Lauren B.

10/07/19

Rebecca M.

would the answer for k be -0.0146?10/07/19