If 18 g of a radioactive substance are present initially and...

If 18 g of a radioactive substance are present initially and 3 yr later only 9.0 g remain, how much of the substance, to the nearest tenth of a gram, will be present after 7 yr? After 7 yr, there will be $\boxed{}$ g of the radioactive substance. (Do not round until the final answer. Then round to the nearest tenth as needed.)

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Transcript

00:01 So in this problem, we're told that there was initially 10 grams of a radioactive substance, and then nine years later, only 5 grams remain.

00:08 We want to know how much of the substance is going to remain after year 14.

00:13 Well, first off, we have to find the rate at which is declining.

00:16 So remember, the amount of the substance will equal to the initial amount times e to the rt power.

00:23 So again, the initial amount in this particular case is 10, and we know that nine years later, there's five grams left.

00:29 So a of t will equal to 5.

00:32 E is a constant.

00:33 We don't know r, but we know nine years later it will equal to nine.

00:37 So of r times nine.

00:39 So now we just have to solve this equation for r at first.

00:42 So we'll start by dividing by 10.

00:44 So 5 divided by 10 is equal to 1 half or 0 .5.

00:47 And this will equal to e to the 9r power.

00:50 So because we're dealing with base e, i'm next going to take the natural log of both sides.

00:55 And the reason why i do this is because one of our properties of logs says if we have, the natural log of e to a power, it's essentially just the exponent.

01:04 So the natural log of e to the r is just 9r.

01:07 So we'll have the natural log of 0 .5 equal to 9r...

Question

Please give Ace some feedback