Video: Simplifying Algebraic Expressions Using Laws of Exponents Involving Rational and Negative Exponents

Simplify ((25)^(3/2 π‘₯) Γ— (8)^(π‘₯ βˆ’ (5/3))/((100)^(3/2 π‘₯) Γ— √(100)).

04:24

Video Transcript

Simplify 25 to the three-halves π‘₯ power times eight to the π‘₯ minus five-thirds power all over 100 to the three-halves π‘₯ power times the square root of 100.

First, we can go ahead and copy down the expression. The first thing I notice is that we can quickly simplify the square root of 100. We can write that as 10. The next thing we notice is that we’re dealing with two square numbers. We have the number 25 and the number 100. We can write 25 as five squared, and then we’ll have five squared to the three-halves π‘₯ power. We can write 100 as 10 squared so that we have 10 squared to the three-halves π‘₯ power. What about eight? Eight is not a square number, but we should recognize that eight is a cubed number. Eight is equal to two cubed. And so eight to the π‘₯ minus five-thirds power is equal to two cubed to the π‘₯ minus five-thirds power.

And then we’ll fill out the rest of the problem. We need to think about a power of a power. π‘Ž to the π‘₯ power to the 𝑦 power is equal to π‘Ž to the π‘₯ times 𝑦 power. And that means five squared to the three-halves π‘₯ power is equal to five to the two times three-halves π‘₯ power. And two times three-halves equals three. So two times three-halves π‘₯ will equal three π‘₯. Following the same procedure for the two cubed, we have two to the three times π‘₯ minus five-thirds power. And we distribute that three so that we have three π‘₯ minus three times five-thirds. But three times five-thirds will equal five.

We’ll follow this process one more time for the 10, which is again two times three-halves π‘₯. And that will simplify to three π‘₯. And then we just bring over our 10. To clean that up, we get five to the three π‘₯ power times two to the three π‘₯ minus five power all over 10 to the three π‘₯ power times 10 to the first power. And again, we’ll need to do some regrouping. We know that π‘Ž to the π‘₯ power times π‘Ž to the 𝑦 power equals π‘Ž to the π‘₯ plus 𝑦 power. And that means if we have something that is π‘Ž to the π‘₯ plus 𝑦 power, we can separate it back into its two parts.

For this, two to the three π‘₯ minus five power, we could rewrite that as two to the three π‘₯ power times two to the negative five power. And then we have five to the three π‘₯ power times two to the three π‘₯ power times two to the negative five power over 10 to the three π‘₯ times 10 to the first. At this point, we should think π‘Ž times 𝑏 to the π‘₯ power is equal to π‘Ž to the π‘₯ power times 𝑏 to the π‘₯ power. And since five and two are both being taken to the three π‘₯ power, we can rewrite this as five times two to the three π‘₯ power. And then our expression becomes five times two to the three π‘₯ power times two to the negative five power all over 10 to the three π‘₯ power times 10 to the first power.

Now, we see where this is going. Five times two equals 10. And we have 10 to the three π‘₯ power in the numerator and 10 to the three π‘₯ power in the denominator. This will equal one. We now have two to the negative five power over 10 to the first power. And π‘Ž to the negative π‘₯ power equals one over π‘Ž to the π‘₯ power. So we have one over two to the fifth power times 10. Two to the fifth power equals 32. So we have 32 times 10 in the denominator. And the expression is equal to one over 320. And so we can say that this expression, simplified and evaluated, is one over 320.

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