1.6 E) Raising a Power to Another Power – Hard
a and b in the equation below are positive integers.
Work out the values of a and b.
On the left side of the equation, we have a power to another power. We can take the expression out of the bracket by multiplying all of the powers inside the bracket by the power that is on the outside of the bracket (3). Before we multiply the powers on the inside of the bracket by the power on the outside of the bracket, I am going to add in a power of 1 for the unknown a; this results in a becoming a1. The adding in of the power of 1 to a is to ensure that I do not forget to multiply any of the powers inside the bracket by the power that is on the outside of the bracket. The adding in of a power of 1 for a results in the equation becoming:
We now multiply all of the powers on the inside of the bracket by the power that is on the outside of the bracket; we multiply all of the powers inside the bracket by 3.
The next step is to equate parts of the expressions on the left and right side of the equation. We can equate the number part of the expressions and the unknown (x) part of the expressions. When we equate the number part of the expressions, we see that a3 and 8 are equal to each other. Also, when we equate the unknown part of the expressions, we see that x3b and x15 are equal to one another. This gives us the two equations that are shown below:
We can now solve these equations to find the values of a and b. Both of these equations only have one unknown in them (ignoring x as an unknown), and this means that we can solve the equations to find the values of a and b in any order.
Let’s solve the number equation first to find the value of a. This equation is shown below:
We want to find the value of a and not a3. Therefore, we cube root both sides of the equation.
This tells us that a is 2.
Let’s now solve the unknown (x) equation to find the value of b. This equation is shown below:
Both sides of the equation have a base of x, and this means that the powers must be the same; 3b must be equal to 15. Therefore, we can get rid of the bases on both sides of the equation to just leave the powers. The equation becomes:
We want to find the value of b and not 3b. Therefore, we divide both sides of the equation by 3.
This tells us that b is 5.
We have now found the values of both a and b; a is 2 and b is 5.
We are now going to look at another example that is very similar to the first example. The only difference with this example is that the unknowns that we are looking for will be in different positions.
c and d in the equation below are positive integers.
Work out the values of c and d.
Like the previous example, the first step in answering this question is to take the expression on the left side of the equation out of the bracket, which we do by multiplying all of the powers on the inside of the bracket by the power that is on the outside of the bracket. Before we multiply the powers, I am going to add in a power of 1 for the 3; the 3 becomes 31.
We now multiply all of the powers on the inside of the bracket by the power that is on the outside of the bracket; we multiply all of the powers inside the bracket by c.
The next step is to equate parts of the expressions on the left and right side of the equation. We can equate the number part of the expressions and the unknown (y) part of the expressions. When we equate the number part of the expressions, we see that 3c and 81 are equal to each other. Also, when we equate the unknown part of the expressions, we see that y6c and yd are equal to one another. This gives us the two equations that are shown below:
The equation that we obtained from equating the unknowns (y) has two unknowns in it; the equation has both c and d in it. The presence of the two unknowns means that we are unable to solve this equation. Therefore, we need to solve the equation that we obtained from equating the numbers first because there is only one unknown in this equation. The equation from the numbers is shown below:
From the above working, we can see that 3 has been multiplied 4 times. This means that c is 4.
We now need to find the value of d, which we do by subbing c as 4 into the equation that we obtained from equating the unknowns (y). The equation from equating the unknowns is shown below:
We now sub in c as 4.
Both of the bases for the two terms above are y. This means that the powers must be the same. Therefore, d must be 24.
We now have the values of both of the unknowns; c is 4 and d is 24.
The final two examples that we are going to look at in this section will involve changing the value of the base for a power. The process used to answer these types of questions is best explained through the examples.
Write 1254a as 5something
We answer these types of questions by modifying the base number of the power. We want to have our answer as 5something, so we modify our current base number (125) so that it becomes 5 to a power.
We can find out what the value of the question mark is by multiplying 5 by itself until we get 125. When we do this, we see that 5 is multiplied 3 times, which means that 53 is 125; we replace 5? with 53.
We now have a power to another power. We can get rid of the brackets by multiplying the powers. The working is shown below:
We now have our expression as 5something, and the value for the something is 12a.
Write 85y x 163y + 2 in the form of 2ay + b
a and b are positive integers
When we multiply powers, the bases of the powers need to be the same. Currently, the base of the first term is 8 and the base of the second term is 16. This means that we cannot multiply the terms in their current form.
The question wants us to give our answer as a power with a base of 2. Therefore, it tells us that we are able to modify both of the terms to get them into the form of 2something. When we have both of the terms in the form 2something, we can multiply the two terms because both of the bases will be 2. We get both of the terms in the form of 2something by using the same process as the previous example.
I am going to place the first term, 85y, into the form of 2something. We are able to do this by changing the base that is currently 8 to become 2 with a power.
We can find out what the value of the question mark is by multiplying 2 by itself until we get 8. When we do this, we see that 23 is 8; we replace 2? with 23.
We now have a power to another power. We can get rid of the brackets by multiplying the powers. The working is shown below:
We now modify the second term, 163y + 2, so that it is in the form of 2something. We are able to do this by changing the base that is currently 16 to become 2 with a power.
We can find out what the value of the question mark is by multiplying 2 by itself until we get 16. When we do this, we see that 24 is 16; we replace 2? with 24.
We now have a power to another power. We can get rid of the brackets by multiplying the powers. The power that is outside the bracket contains two terms. Therefore, when you are multiplying powers, you need to make sure that the power on the inside of the bracket is multiplied by both of the terms in the power on the outside of the bracket. The working is shown below:
We now have the first and second terms as 2something. The calculation with both terms as 2something is:
Both of the bases for the terms are 2, and this means that we can multiply the terms. When we multiply powers with the same base, we add the powers. The working is shown below.
We now have our final answer in the form of 2ay + b. The final step is to say what a and b are; a is 27 and b is 8.