Interpreting change in exponential models: with manipulation | High School Math | Khan Academy

Ocean sunfishes are well known for rapidly gaining a lot of weight on a diet based on jellyfish. The relationship between the elapsed time ( t ) in days since an ocean sunfish is born and its mass ( m(t) ) in milligrams is modeled by the following function.

All right, complete the following sentence about the daily percent change in the mass of the sunfish: Every day there is a blank percent addition or removal from the mass of the sunfish.

So one thing that we can, we know from almost from the get-go, we know that the sunfish gains weight. We also see that as ( t ) grows, as ( t ) grows, the exponent here is going to grow. If you grow an exponent on something that is larger than one, ( m(t) ) is going to grow.

So I already know it's going to be about addition to the mass of the sunfish. But let's think about how much is added every day. Let's think about it. Well, let's see if we can rewrite this. This is—I'm going to just focus on the right-hand side of this expression, so ( 1.35^{t/6} + 5 ). That's the same thing as ( 1.35^5 \times 1.35^{t/6} ), and that's going to be equal to ( 1.35^5 \times 1.35 ).

I can separate this ( t/6 ) as ( \frac{1}{6} \times t ), so ( 1.35^{1/6} ) and then that being raised to the ( t ) power. So let's think about it. Every day as ( t ) increases by 1, now we can say that we're going to take the previous day's mass and multiply it by this common ratio.

The common ratio here isn't the way I've written it; it isn't ( 1.35 ), it's ( 1.35^{1/6} ). Let me draw a little table here to make that really, really clear. All of that algebraic manipulation I just did is just so I could simplify this.

So I have some common ratio to the ( t ) power. Based on how I've just written it, when ( t ) is zero, well, as ( t ) is zero, this is one. So then we just have our initial amount; our initial mass is going to be ( 1.35^5 ).

And then when ( t ) is equal to 1, when ( t ) is equal to 1, it's going to be our initial mass ( 1.35^5 ) times our common ratio times ( 1.35^{1/6} ). When ( t ) equals 2, we're just going to multiply what we had at ( t ) equals 1 and we're just going to multiply that times ( 1.35^{1/6} ) again.

And so every day—well let me get—every day we are growing. Every day we're growing by our common ratio ( 1.35^{1/6} ). Actually, let me get a calculator out; we're allowed to use calculators in this exercise.

So ( 1.35^{(1/6)} ) power is equal to approximately 1.051. So this is approximately ( 1.35 \times 1.051^t ).

Well, growing by a factor of 1.051 means that you're adding a little bit more than five percent. You're adding 0.51 every day of your mass. So that's—you're adding 5.1. And if you're rounding to the nearest percent, we would say there is a five percent addition to the mass of the sunfish every day.