Video: Solving a System of Exponential Equations

Given that 2^(π‘š) Γ— ^(𝑛) = 2048 and 2^(𝑛) Γ— 4^(π‘š) = 1024, determine the values of π‘š and 𝑛.

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Video Transcript

Given that two to the power of π‘š multiplied by four to the power of 𝑛 equals 2048 and two to the power of 𝑛 multiplied by four to the power of π‘š is equal to 1024, determine the values of π‘š and 𝑛.

Well, the first thing we want to do is rewrite it so that each of our terms is expressed as in powers of two. So, when we do that, we’ve got two to the power of π‘š multiplied by two squared to the power of 𝑛 equals 2048. And we’ve got two to the power of 𝑛 multiplied by two squared to the power of π‘š equals 1024.

Well, it’s here where we can apply one of our exponent laws. And that’s if we have π‘₯ to the power of π‘Ž all to the power of 𝑏, this is gonna be equal to π‘₯ to the power of π‘Žπ‘. So, we can multiply the exponents. So, in both of our examples, it means that we can multiply two and 𝑛 and two and π‘š.

So, we’re gonna start with the left-hand side first. So, we’ve got two to the power of π‘š is equal to two to the power of two 𝑛, because we multiplied the two and the 𝑛 together, is equal to 2048. Well, therefore, we’ve got two to the power of π‘š multiplied by two to the power of two 𝑛 is equal to two to the power of 11. And that’s because 2048 is two to the power of 11. But as we have the same base throughout, what we can do is we can equate the exponents.

And to enable us to do that, we’re gonna use the second exponent law. And this is that if we have π‘₯ to the power of π‘Ž multiplied by π‘₯ to the power of 𝑏, then we have π‘₯ to the power of π‘Ž plus 𝑏. So, we add the exponents, if we got the same base and we multiply them together. So, in our case, we’re gonna get π‘š plus two 𝑛 is equal to 11. And that’s because we had two to the power of π‘š multiplied by two to power of two 𝑛 is equal to two to the power of 11.

So, now we’re gonna do the same to the right-hand side. So, first of all, we get two to the power of 𝑛 multiplied by two to the power of two π‘š is equal to 1024. So therefore, we’re gonna have two to the power of 𝑛 multiplied by two to the power of two π‘š is equal to two to the power of 10. That’s because 1024 is two to the power of 10. And we could see that because if we divided 2048 by two, we get 1024.

And, once again, we’ve got the same base, so what we can do is we can equate the exponents. And we do that using the second exponent law again. And therefore, when we do that, we get 𝑛 plus two π‘š is equal to 10. So, now we have a pair of simultaneous equations because we’ve got two unknowns, that’s π‘š and 𝑛. And we’re gonna solve these to find π‘š and 𝑛.

And the first thing we’re gonna do is we’re gonna subtract two π‘š from both sides of equation two. And the reason we’re gonna do that is we’re gonna solve this using the substitution method. And to solve it using the substitution method, what we need to do is to change the subject so that either π‘š or 𝑛 are the subject of the equation.

So, I’ve done that in this case and I’ve made 𝑛 the subject by subtracting two π‘š from each side. So, I’ve got 𝑛 equals 10 minus two π‘š. And we’re gonna call that equation three. I’ve labelled each of the equations because it makes it easier to see what we’re doing when we do it step-by-step.

So, now the next step is to substitute equation three into equation one. So, when we do that, we’re gonna get π‘š plus two multiplied by then we’ve got 10 minus two π‘š. That’s because 𝑛 was equal to 10 minus two π‘š. And then this is equal to 11. And then, when we expand the parentheses, we get π‘š plus 20 minus four π‘š equals 11. And then, when we tidy this up, we get negative three π‘š plus 20 is equal to 11. And then, what we’re gonna do is solve this to find 𝑛.

So, we’re gonna add three π‘š to each side of the equation and subtract 11. And that’s because we want positive π‘šs. And when we do that, we get nine is equal to three π‘š. And that’s because if you add three π‘š onto both sides of the equation, we get three π‘š on the right-hand side. And if we subtract 11, well, 11 away from 20 gives us nine. So, nine equals three π‘š. And then, if we divide both sides of the equation by three, we get π‘š is equal to three. So, that’s the first part solved. Now what we need to do is to find 𝑛.

Well, to find 𝑛, what we need to do is substitute π‘š equals three into equation three. So, we’re gonna get 𝑛 is equal to 10 minus then you’ve got two multiplied by three cause we had 𝑛 is equal to 10 minus two π‘š. So therefore, we get 𝑛 is equal to four. And that’s because we had 10 minus six which gives us four.

So, we’ve solved the problem. And we’ve found the values of π‘š and 𝑛. But what we’re gonna do now is just quickly check it. Well, to check it, I’m gonna substitute our π‘š- and 𝑛-values back in.

So, we’re gonna have two to the power of three multiplied by four to the power of four. So, we’re gonna do the first equation we had. Well, this is gonna give us eight multiplied by 256. And that’s cause two cubed is eight. And four to the power of four is 256, which gives us an answer of 2048, which is the same as we were given in the question. So therefore, we can say that, given that two to the power of π‘š multiplied by four to the power of 𝑛 is equal to 2048 and two to the power of 𝑛 multiplied by four to the power of π‘š is equal to 1024, the values of π‘š and 𝑛 are three and four, respectively.

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