Write an exponential equation in the form 𝑦 is equal to 𝑎 multiplied by 𝑏 to the power of 𝑥 for the numbers in the table.
We are given three pairs of values. When 𝑥 equals one, 𝑦 equals 15; when 𝑥 equals two, 𝑦 is equal to 25; and when 𝑥 is three, 𝑦 is equal to 125 over three. We need to calculate the values of the constants 𝑎 and 𝑏 by substituting our pairs into the equation. When 𝑥 is equal to one and 𝑦 is equal to 15, we have 15 is equal to 𝑎 multiplied by 𝑏 to the power of one. As anything to the power of one is equal to itself, we are left with 15 is equal to 𝑎 multiplied by 𝑏. We will call 15 equals 𝑎𝑏 equation one.
Substituting our second pair of values gives us 25 is equal to 𝑎 multiplied by 𝑏 squared. This can be rewritten as 25 equals 𝑎𝑏 squared. And we will call this equation two. We now have a pair of simultaneous equations which we need to solve to calculate 𝑎 and 𝑏. One way to solve these is to divide equation two by equation one. This is known as elimination. On the left-hand side, we have 25 over 15 and on the right-hand side 𝑎𝑏 squared divided by 𝑎𝑏. We can cancel an 𝑎 and a 𝑏 on the right-hand side. 25 and 15 are both divisible by five; therefore, 𝑏 is equal to five-thirds.
We can then substitute this value back into equation one or equation two to calculate 𝑎. We will use equation one. This gives us 15 is equal to 𝑎 multiplied by five-thirds. Dividing both sides by five-thirds gives us 𝑎 is equal to nine. We recall that when dividing by a fraction, we can multiply by the reciprocal. 15 divided by five-thirds is the same as 15 multiplied by three-fifths. As one-fifth of 15 is three, three-fifths will be equal to nine. We now have values of 𝑎 and 𝑏 that we can substitute back in to our equation. The exponential equation for the numbers in the table is 𝑦 is equal to nine multiplied by five-thirds to the power of 𝑥.
At this stage it is also worth checking that this equation holds for the third pair of numbers in the table. Clearing some space and substituting in our values, we have 125 over three is equal to nine multiplied by five-thirds or five over three cubed. When cubing any fraction, we can keep the numerator and denominator separately. So we are left with nine multiplied by five cubed over three cubed. Five cubed is 125 and three cubed is 27. We need to multiply nine by 125 over 27. Nine and 27 are divisible by nine. This means we are left with 125 over three, which means that the equation 𝑦 is equal to nine multiplied by five-thirds to the power of 𝑥 is true for the third pair of values.