Question Video: ο»ΏUsing Substitution of Terms to Verify an Identity Mathematics

Let 𝑙 = (1/2)(6^(π‘₯) + 6^(βˆ’π‘₯)) and π‘š = (1/2)(6^(π‘₯) + 6^(βˆ’π‘₯)). Is it true that 𝑙² βˆ’ π‘šΒ² = 1?

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Video Transcript

Let 𝑙 equal one-half times six to the π‘₯ power plus six to the negative π‘₯ power and π‘š equal one-half times six to the π‘₯ power minus six to the negative π‘₯ power. Is it true that 𝑙 squared minus π‘š squared equals one?

If we plug in what we know for 𝑙 and π‘š, we get the expression one-half times six to the π‘₯ power plus six to the negative π‘₯ power squared minus one-half times six to the π‘₯ power minus six to the negative π‘₯ power squared. Because we’re going to have to be squaring some of these terms, I notice that we have six to the π‘₯ power and six to the negative π‘₯ power in both the 𝑙- and the π‘š-variable.

One strategy to make our calculations a bit simpler would be to let π‘Ž equal six to the π‘₯ power and 𝑏 equal six to the negative π‘₯ power. If we do that, we end up with one-half times π‘Ž plus 𝑏 squared minus one-half times π‘Ž minus 𝑏 squared. We distribute our square to the one-half and the π‘Ž plus 𝑏 term and to the one-half and π‘Ž minus 𝑏 term. One-half squared is one-fourth. And we also know the form for π‘Ž plus 𝑏 squared. It’s equal to π‘Ž squared plus two π‘Žπ‘ plus 𝑏 squared. So we’ll add that. Bring down the subtraction. One-half squared is one-fourth.

And we have to be careful here because π‘Ž minus 𝑏 squared is not the difference of squares. π‘Ž minus 𝑏 squared will be equal to π‘Ž squared minus two π‘Žπ‘ plus 𝑏 squared. So we substitute π‘Ž squared minus two π‘Žπ‘ plus 𝑏 squared back into our equation. At this point, we have a one-fourth term that we could remove. We are undistributing that one-fourth. So we’ll have one-fourth multiplied by π‘Ž squared plus two π‘Žπ‘ plus 𝑏 squared. And then we’re subtracting this entire term, which means we’ll have negative π‘Ž squared. Subtracting negative two π‘Žπ‘ means we’ll be adding two π‘Žπ‘, and then subtracting a positive 𝑏 squared. So we have minus 𝑏 squared.

Before we do anything else, we wanna check and see if we can simplify by combining like terms. We have a positive π‘Ž squared term and a negative π‘Ž squared term, which cancels out. We have a positive 𝑏 squared term and a negative 𝑏 squared term, which also cancels out. And then we have two π‘Žπ‘ plus two π‘Žπ‘. We combine them by combining their coefficients so that we have four π‘Žπ‘. And if four π‘Žπ‘ is multiplied by one-fourth, one-fourth times four equals one. And we’re left with π‘Ž times 𝑏.

Now we need to go back in and plug in the values we know for π‘Ž and 𝑏. We’ll multiply six to the π‘₯ power times six to the negative π‘₯ power. To do that, we’ll have to remember our exponent rules when we have the same base taken to different powers being multiplied together. π‘₯ to the π‘Ž power times π‘₯ to the 𝑏 power is equal to π‘₯ to the π‘Ž plus 𝑏 power. Since both of these exponents have a base of six, we’ll combine their powers, which will be π‘₯ plus negative π‘₯. π‘₯ plus negative π‘₯ equals zero, and six to the zero power equals one.

Now, if we go back and look at our original question, we want to know is 𝑙 squared minus π‘š squared equal to one. We began with 𝑙 squared minus π‘š squared, which gave us π‘Ž times 𝑏. And then we found that π‘Ž times 𝑏 was equal to one. And so we can say, under these conditions, yes, 𝑙 squared minus π‘š squared is equal to one.

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