Question Video: Creating Exponential Equations with One Variable to Solve Problems | Nagwa Question Video: Creating Exponential Equations with One Variable to Solve Problems | Nagwa

Question Video: Creating Exponential Equations with One Variable to Solve Problems Mathematics • Second Year of Secondary School

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A cereal manufacturer decides to make their products healthier by reducing the amount of sugar in them. Their target is to reduce the amount of sugar in their product range by 20%. They plan to achieve their target in 4 years. Write an equation they could use to find π‘Ÿ, the annual sugar reduction rate required, to achieve their target.

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Video Transcript

A cereal manufacturer decides to make their products healthier by reducing the amount of sugar in them. Their target is to reduce the amount of sugar in their product range by 20 percent. They plan to achieve their target in four years. Write an equation they could use to find π‘Ÿ, the annual sugar reduction rate required, to achieve their target.

We’re asked to write an equation that could be used by a cereal manufacturer to find the reduction rate π‘Ÿ required to achieve a 20 percent reduction in sugar in their products over four years. To produce the required equation, we first recall that an exponential decrease or decay can be represented by the function 𝑦 of 𝑑 is equal to π‘Ž multiplied by 𝑏 raised to the power 𝑑 for zero less than 𝑏 less than one. This is equivalent to π‘Ž multiplied by one plus capital 𝑅 raised to the power 𝑑. And that’s for 𝑅 less than zero. And this is where π‘Ž is equal to 𝑦 at 𝑑 is equal to zero; that’s the initial value. 𝑏 is equal to one plus capital 𝑅. And 𝑅 is the constant rate of change of the function 𝑦. And it’s this form that we’ll be concerned with.

Now, since we know that we want the quantity of sugar to decrease, we can represent the decay rate as a percentage. That’s uppercase 𝑅, capital 𝑅, is equal to negative lowercase π‘Ÿ over 100. So, our reduction rate is π‘Ÿ percent. And our exponential decrease or decay function can be rewritten as 𝑦 of 𝑑 is equal to π‘Ž multiplied by one minus π‘Ÿ over 100 raised to the power 𝑑.

Now, using some of the information provided in the question, we’re told that the target is a 20 percent reduction in sugar in four years. So, if we let 𝑦 be the amount of sugar, then we have 𝑦 at 𝑑 is equal to four is equal to 0.8 times π‘Ž, that is, 80 percent of the initial amount π‘Ž. But we can also substitute 𝑑 is equal to four into our exponential decay function 𝑦 of 𝑑. And this gives us 𝑦 at 𝑑 is equal to four is equal to π‘Ž multiplied by one minus π‘Ÿ over 100 raised to the power of four. And now equating this with our equation above, we have π‘Ž multiplied by one minus π‘Ÿ over 100 raised to the power of four is equal to 0.8 multiplied by π‘Ž. Now dividing both sides by π‘Ž, we have one minus π‘Ÿ over 100 raised to the power of four is equal to 0.8.

Now in our parentheses, if we put everything over the common denominator of 100, we have 100 minus π‘Ÿ over 100 raised to the power of four is equal to 0.8. The cereal manufacturer could therefore use the equation 100 minus π‘Ÿ over 100 all raised to the power of four is equal to 0.8 to find the value of π‘Ÿ, that is, the annual sugar reduction rate required to achieve their target of a 20 percent reduction in four years. And of course, this equation can be rearranged to make π‘Ÿ the subject.

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