Video: Writing and Evaluating Exponential Functions to Model Exponential Decay in a Real-World Context

The production of a gold mine is decreasing with a rate of 7% annually. Given that the mine’s production was 6400 kg in the first year, determine its production in the sixth year giving your answer to the nearest integer.

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Video Transcript

The production of a gold mine is decreasing with a rate of 7 percent annually. Given that the mine’s production was 6400 kilograms in the first year, determine its production in the sixth year giving your answer to the nearest integer.

We have to find the production in the sixth year. We’re told what the production was in the first year and also how the production is changing each year. It’s going down by 7 percent year-on-year. Let’s start a table of values. In year one, the production was 6400 kilograms.

We want to find out what happened in year six. But we’ll work our way up to that. Let’s first ask what happens in year two. As production is decreasing with a rate of 7 percent annually, this value should be the 6400 kilograms of the year before minus 7 percent of that 6400. Another way of saying 7 percent of 6400 is 0.07 times 600 [6400].

This is now something that we could put in our calculator. But let’s go a step further. We can write 6400 as one times 6400. And then we can factor out the common 6400 of these terms to get 0.93 times 6400. We’ll put this in the table without evaluating it. This will help us to see a pattern later.

We see that as the production of the gold mine is decreasing with a rate of 7 percent annually, the production in the second year is 0.93 times the production in the first year. Another way of saying this is that this 7 percent decrease means that we left with 93 percent.

What happened in year three? Well there’s another 7-percent decrease. This decrease happens annually. And so the production in year three is 93 percent of the production in year two. This is 0.93 times 0.93 times 6400 kilograms. We can rewrite 0.93 times 0.93 as 0.93 squared or 0.93 to the power of two, like so.

In year four, the production is decreased again by 7 percent. And again we can combine the new factor of 0.93 with the others. We’re hopefully starting to see a pattern emerging. In year five we’re multiplying by 0.93 again to get 0.93 to the power of four times 6400. And by multiplying by 0.93 one last time, we get the production in year six, which is what the question asks us for.

Before we go about evaluating the answer, let’s first note the pattern. In all the years since year three, the exponent of 0.93 has been one less than the year. We can actually continue this pattern back further by writing 0.93 as 0.93 to the power of one. And we can actually go back even further. You might know that’s 0.93 the power of zero is one. And so multiplying by 0.93 to the power of zero does nothing to the value of 6400.

We could express this relationship using the formula production equals 0.93 to the power of year minus one times 6400. Substituting in a value of six for the year will give the production in the sixth year. And this will agree with the value that we found in the last row of our table.

The answer is 0.93 to the power of five times 6400 kilograms which, using a calculator, we find this to be 4452.405 dot dot dot kilograms. Rounding to the nearest integer as we’re told to, we get that the production in the sixth year is 4452 kilograms.

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