Question Video: Converting Rational Exponents into Radical Form | Nagwa Question Video: Converting Rational Exponents into Radical Form | Nagwa

Question Video: Converting Rational Exponents into Radical Form Mathematics

Express π‘₯^(16/3) Γ— π‘₯^(23/2) in the form the π‘šth root of π‘Ž^𝑛.

02:25

Video Transcript

Express π‘₯ to the power of 16 over three multiplied by π‘₯ to the power of 23 over two in the form the π‘šth root of π‘Ž to the power of 𝑛.

To solve this problem, we have to write some rules of exponents that’ll help us out. So if we’re gonna express π‘₯ to the power of 16 over three multiplied by π‘₯ to the power of 23 over two, the first rule we’re gonna use is this one. Our first rule we are looking at is that π‘Ž to the power of π‘š multiplied by π‘Ž to the power of 𝑛 is equal to π‘Ž to the power of π‘š plus 𝑛.

So let’s apply this to our question; that’s gonna give us π‘₯ to the power of 16 over three plus 23 over two. And to add these powers, we’re gonna make the denominators equal. So we’re gonna have π‘₯ to the power of- we get 23 over six. And we get that by multiplying the numerator and the denominator both by two. And we do that because six is the lowest common multiple of two and three, which are both for our denominators.

And then we’re gonna have plus 69 over six, again this time multiplying the numerators on it by three to give us that denominator of six. And then when we add the powers together, we get our final answer, which is π‘₯ to the power of 101 over six. But is this it? Is this the problem solved? That’s actually a no because having a look at the original question, we can see that we need to leave it in this particular form. As to be able to do this, what we’re gonna do is actually use another rule of exponents. And this rule of exponents shows us that π‘Ž to the power of 𝑛 over π‘š is equal to the π‘šth root of π‘Ž to the power of 𝑛.

Okay, great! Let’s apply this to the term we have. Well first of all, we know that is going to be sixth root because if we look at our rule of exponents we have that the bottom of the denominator, so the π‘š in this case, is equal to six in our term. And then our power of π‘₯ is gonna be 101 because again looking at this, this is the numerator, so our 𝑛 value if you look at the rule of exponents. It’s there we have it: π‘₯ to the power of 16 over three multiplied by π‘₯ to the power of 23 over two is equal to the sixth root of π‘₯ to the power of 101.

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