Question Video: Simplifying Algebraic Expressions Using Laws of Exponents | Nagwa Question Video: Simplifying Algebraic Expressions Using Laws of Exponents | Nagwa

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

Join Nagwa Classes

Attend live General Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

Simplify ((36)^(π‘₯ + (1/2) Γ— (8)^(π‘₯ + 1))/((6)^(π‘₯ βˆ’ 1) Γ— (12)^(π‘₯ + 5) Γ— (√16)^(π‘₯ βˆ’ 1)).

07:46

Video Transcript

Simplify 36 raised to the power of π‘₯ plus one-half multiplied by eight to the power of π‘₯ plus one all divided by six to the power of π‘₯ minus one multiplied by 12 to the power of π‘₯ plus five multiplied by the square root of 16 raised to the power of π‘₯ minus one.

In this question, we’re asked to simplify an expression involving exponents. And there’s a few different ways we could approach this. Instead of trying to simplify the entire expression at once, we’re going to simplify each part separately.

Let’s start with the first factor in the numerator: 36 to the power of π‘₯ plus one-half. We can see this is an exponential expression. It’s a base raised to an exponent. And in this case, the base is an integer; it’s 36. And in this expression, we can see there are many integer-based exponential expressions. And when simplifying exponential expressions in this form, it’s usually a good idea to find the prime factorization of the base, since this will help us simplify the expression later.

In this case, 36 has two factors of two and two factors of three; 36 is two squared times three squared. Therefore, we can write 36 to the power of π‘₯ plus one-half as two squared times three squared all raised to the power of π‘₯ plus one-half.

We can see that in our base we have a product and in our exponent we have a sum. This gives us two methods for simplifying this expression. We can use either of the product rules for exponents. We’re going to use the fact that we can distribute the exponent over a product. π‘Ž times 𝑏 all raised to the 𝑛th power is π‘Ž to the 𝑛th power multiplied by 𝑏 to the 𝑛th power. In this case, our exponent 𝑛 is π‘₯ plus one-half. So we get two squared all raised to the power of π‘₯ plus one-half multiplied by three squared raised to the power of π‘₯ plus one-half.

And now, each of these factors is in the form π‘Ž to the power of 𝑛 all raised to the power of π‘š, which we recall is equal to π‘Ž to the power of 𝑛 times π‘š. Therefore, we can use this to simplify our expression. And we note that two multiplied by π‘₯ plus one-half is equal to two π‘₯ plus one. Therefore, we have two to the power of two π‘₯ plus one multiplied by three to the power of two π‘₯ plus one.

Let’s now move on to simplifying the second factor in our numerator. To do this, let’s clear some space and follow the same process. We’ll start by factoring the base eight into primes. And we can do this by noting that eight is two cubed and two is a prime. This means we can rewrite eight as two cubed, giving us two cubed all raised to the power of π‘₯ plus one. And now we can see this is in the form π‘Ž to the power of 𝑛 all raised to the power of π‘š, which we know is equal to π‘Ž to the power of 𝑛 times π‘š. Therefore, we can simplify this further by multiplying the exponents. Three times π‘₯ plus one is three π‘₯ plus three. So we get two to the power of three π‘₯ plus three.

Let’s now move on to simplifying the first factor in the denominator. We’ll clear some space and do this by following the same process. First, we’ll factor six into primes. It’s equal to two times three. This gives us two times three all raised to the power of π‘₯ minus one. To simplify this further, we want to use one of our laws of exponents. And to decide which rule we want to use, we need to note that we want to write our expression in terms of bases being prime numbers. That’s two and three. And we can do this by distributing the exponent over each of the factors. π‘Ž times 𝑏 all raised to the 𝑛th power is equal to π‘Ž to the 𝑛th power multiplied by 𝑏 to the 𝑛th power. Applying this, we get two to the power of π‘₯ minus one multiplied by three to the power of π‘₯ minus one.

We can follow the same process to simplify the second factor in our denominator. We start by factoring 12 into primes. 12 is two times six, so we get an extra factor of two. It’s two squared times three. Once again, we’re going to distribute the exponent over each of the factors. This then gives us two squared to the power of π‘₯ plus five multiplied by three to the power of π‘₯ plus five. And we can simplify this further. The first factor in this expression is two squared all raised to the exponent π‘₯ plus five.

And we can recall if we raise π‘Ž to the power of 𝑛 and then raise all of this to the power of π‘š, it’s equal to π‘Ž raised to the product 𝑛 times π‘š. Therefore, since two multiplied by π‘₯ plus five is equal to two π‘₯ plus 10, we simplify this expression to be two to the power of two π‘₯ plus 10 multiplied by three to the power of π‘₯ plus five.

We now need to simplify the final factor in the denominator. To do this, let’s start by clearing some space. We’re going to first want to rewrite the base as an integer. We can do this by just evaluating the base. We recall the square root of 16 is the nonnegative number whose square is equal to 16. So, four squared is equal to 16. So, we know the square root of 16 is equal to four. Therefore, root 16 to the power of π‘₯ minus one is four to the power of π‘₯ minus one.

And now we can follow the same process. We factor four into primes; it’s two squared. This gives us two squared to the power of π‘₯ minus one. And we can simplify this by using the same result. π‘Ž to the power of 𝑛 all raised to the power of π‘š is π‘Ž to the power of 𝑛 times π‘š. And then, since two multiplied by π‘₯ minus one is two π‘₯ minus two, this expression simplifies to give us two to the power of two π‘₯ minus two.

And now that we’ve written each of the five parts of the expression we’re given in the question in this form, we can start simplifying. We’ll start by clearing some space and then substituting each of the five expressions we found into this expression. And doing this gives us the following. We can now see that this is the product and quotient of exponential functions where the base is either two or three.

And now we can just simplify this expression by using the laws of exponents. First, we recall if we multiply two exponential expressions with the same base, we add their exponents. π‘Ž to the power of 𝑛 times π‘Ž to the power of π‘š is equal to π‘Ž to the power of 𝑛 plus π‘š. And we can apply this separately in the numerator and denominator. Let’s start with the numerator. We have two to the power of two π‘₯ plus one multiplied by two to the power of three π‘₯ plus three.

To multiply these two factors together, we need to add their exponents because they have the same base. This gives us a new exponent of five π‘₯ plus four. Therefore, we’ve rewritten the numerator of this expression as two to the power of five π‘₯ plus four multiplied by three to the power of two π‘₯ plus one. We can then do the same in the denominator. This time, we see we have three factors each with a base of two. So to multiply these together, we need to add their exponents together. This gives us a new exponent of five π‘₯ plus seven.

So, if we simplify these terms, in our denominator we’ll get a factor of two to the power of five π‘₯ plus seven. However, we’re not done yet because in our denominator we also have two factors with a base of three. And since their base is the same, we can multiply these in the same way. We add their exponents, giving us a new exponent of two π‘₯ plus four. This gives us two to the power of five π‘₯ plus four multiplied by three to the power of two π‘₯ plus one all divided by two to the power of five π‘₯ plus seven multiplied by three to the power of two π‘₯ plus four.

And we can simplify this expression even further by noticing we’re taking the quotient of exponential expressions with the same base. And we know when we do this, we take the difference in their exponents. π‘Ž to the power of 𝑛 divided by π‘Ž to the power of π‘š is equal to π‘Ž to the power of 𝑛 minus π‘š. This means there’s two different ways we could simplify this expression. We could cancel shared factors of two and three in the numerator and denominator, or we can use this law of exponents.

Both methods are valid. However, we’re going to use the law of exponents. Taking the difference in the exponents with base two, we have negative three. Similarly, taking the difference in the exponents with base three, we have negative three. Therefore, this entire expression simplifies to give us two to the power of negative three multiplied by three to the power of negative three. And finally, we can evaluate this expression by recalling π‘Ž to the power of negative 𝑛 is equal to one divided by π‘Ž to the power of 𝑛.

Therefore, this is equal to one over two cubed multiplied by one over three cubed. And we can evaluate two cubed and three cubed. We can calculate two cubed is equal to eight and three cubed is 27. And eight times 27 is 216. Therefore, we were able to simplify the expression given to us in the question. We showed that it was equal to one divided by 216.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy