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

Question Video: Simplifying Algebraic Expressions Using Laws of Exponents Involving Rational and Negative Exponents Mathematics • Second Year of Secondary School

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Determine the simplest form of ((16)^((3/2)π‘₯) Γ— 27^(π‘₯ βˆ’ (1/3)))/((144)^((3/2)π‘₯) Γ— √(81)).

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Video Transcript

Determine the simplest form of 16 raised to the power three over two π‘₯ multiplied by 27 raised to the power π‘₯ minus one over three all divided by 144 raised to the power three over two π‘₯ multiplied by the square root of 81.

We’re given an exponential expression in the form of a fraction where both the numerator and denominator consist of products of exponential expressions themselves. To simplify this, we’ll use the laws of exponents together with prime factorization of the bases, recalling that the base π‘Ž is a positive integer not equal to one raised to an exponent 𝑏. We’re going to look at our expression term by term. So let’s first look at the square root of 81 in the denominator. We know that nine squared is 81 so that in fact nine is equal to the square root of 81. And so, we can replace the square root of 81 in our denominator with nine.

Next, let’s break our bases 16, 27, 144, and nine into their prime factors. We know that 16 is four squared, which is two multiplied by itself four times, so that 16 is two raised to the power four. 27 is nine times three, which is three multiplied by itself three times. And that’s three cubed, three raised to the power three. 144 is 12 squared, that is, four times three squared. And that’s two to the power four multiplied by three squared and then nine, which is three squared, is three times three. Making some space, we can put these into our expression.

So our expression is now two raised to the power of four all raised to the power of three over two π‘₯ multiplied by three raised to the power three all raised to the power π‘₯ minus one over three divided by three squared multiplied by two raised to the power of four all to the power three over two π‘₯ multiplied by three squared.

So now let’s consider the first term in our denominator. To break this down, we can use the power of a product rule for exponents. This tells us that π‘Ž multiplied by 𝑏 all raised to the power π‘š is equal to π‘Ž raised to the power π‘š multiplied by 𝑏 raised to the power π‘š. This gives us three squared raised to the power three over two π‘₯ multiplied by two to the fourth power raised to the power three over two π‘₯.

So now if we separate these out in our denominator, we then see that we have a common factor in our numerator and denominator of two raised to the power four all raised to the power three over two π‘₯. And we can then divide through top and bottom by this factor. We’re now left with three to the power of three all raised to the power π‘₯ minus one over three divided by three squared to the power three over two π‘₯ multiplied by three squared.

So all of our bases are now some power of three. And we can use the power law for exponents to simplify each term. The exponent in our numerator then becomes three multiplied by π‘₯ minus one over three, that is, three π‘₯ minus one. And the exponent in the left hand of our denominator becomes two multiplied by three over two π‘₯, that is, three π‘₯. Our numerator is then three raised to the power three π‘₯ minus one and our denominator three raised to the power three π‘₯ multiplied by three squared.

So making some space, we have three raised to the power three π‘₯ minus one divided by three raised to the power three π‘₯ multiplied by three squared. Now in the numerator, we can use the product law for exponents. That is, π‘Ž raised to the power π‘š plus 𝑛 is equal to π‘Ž raised to the power π‘š multiplied by π‘Ž raised to the power 𝑛. In our case, the base π‘Ž is equal to three, the first exponent π‘š is three π‘₯, and the second exponent 𝑛 is negative one.

And so, our numerator becomes three raised to the power three π‘₯ multiplied by three raised to the power negative one. Now we see again that we have a common factor in our numerator and denominator of three raised to the power three π‘₯. And so dividing both numerator and denominator by three raised to the power three π‘₯, we’re left with three raised to the power negative one over three squared.

And so, finally, using the law for negative exponents, that is π‘Ž raised to the power negative π‘š is equal to one over π‘Ž raised to the power π‘š, where in our case π‘š is equal to one. And so, we have one over three squared multiplied by three raised to the power one. And so, in our denominator, we can use the product rule again. And this gives us one over three raised to the power two plus one. That is one over three cubed, which is one over 27.

Therefore, the simplest form of the given expression is one over 27.

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