A9 gram sample of a substance that's used for drug research has a k-value of 0.1483.

n=n0e^-kt

n0 = initial mass (at time t=0)

n = mass at time t

k = a positive constant that depends on the substance itself and on the units used to measure time

t = time, in days.

*find the substance's half-life, in days.
round your answer to the nearest tenth.

So the formula for exponential decay is N=N0e^-kt

9 is plugged into the initial value which is N0.
0.1483 is plugged into the k exponent.
and 4.5 is plugged into N because it is half of 9.

so now we have 4.5=9e^-0.1483t

First we need to get 9 on the other side. so divide both sides by 9.

4.5/9=9/9e^-0.1483t

we end up with 0.5=e^-0.1483t

In order to get the answer we need to do a Log Notation. It is the Ln button in your calculator. I like to use Photomath because of its calculator. So use that if you can.

Now we have Ln(0.5)=Ln(e^-0.1483t)

Don't do that in the calculator just yet though.

Simply looking at it we can tell that we now have Ln(0.5)=-0.1483t because Ln got rid of e and the exponents for us.

Now we need to get rid of -0.1483
Obviously were going to divide both sides by -0.1483

Ln(0.5)/-0.1483=-0.1483t/-0.1483

we get Ln(0.5)/-0.1483=t

Now, put that into your calculator and you should get 4.67395

Since you need to the nearest 10th, round 6 up because of the 7 in the hundreds place.

your answer is t=4.7 or just 4.7

Jjutsclopyrsvnkiu it’s c