{"id":12468,"date":"2020-09-23T02:13:53","date_gmt":"2020-09-22T18:13:53","guid":{"rendered":"https:\/\/www.intmath.com\/blog\/?p=12468"},"modified":"2020-09-23T02:13:53","modified_gmt":"2020-09-22T18:13:53","slug":"how-to-multiply-x-with-different-exponents","status":"publish","type":"post","link":"https:\/\/www.intmath.com\/blog\/mathematics\/how-to-multiply-x-with-different-exponents-12468","title":{"rendered":"How To Multiply X with Different Exponents"},"content":{"rendered":"
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Do you still remember the concept of variables and exponents? Before we jump into the topic in question, let's review.<\/p>\n

x<\/em>3<\/p>\n

x<\/em>\u00a0is a\u00a0variable,<\/strong>\u00a0or something that has an unknown value. The value of\u00a0x\u00a0<\/em>may vary\u00a0(hence the name). In one equation,\u00a0x\u00a0<\/em>may be equal to 4. In another equation,\u00a0x\u00a0<\/em>may be equal to 7 or a more complicated number like 75\/2. We express a variable with a symbol. The most common symbols used in most equations are\u00a0x<\/em>,\u00a0y<\/em>, and\u00a0z<\/em>.<\/p>\n

On the other hand, the number 3 above \"x\"<\/em>\u00a0in \"x<\/em>3\" is what people call an\u00a0exponent<\/strong>. The exponent tells us how many times it uses a number or variable in multiplication. The number or variable we multiply\u00a0is called the\u00a0base<\/strong>.<\/p>\n

Let's use a simple\u00a0example using an integer instead of variable for the base. Do you know the value of 43?<\/p>\n

43 means 4 times 4 times 4. The number 4 is raised to the third power\u00a0or cubed.<\/p>\n

43 \u00a0= 4 * 4 * 4 = 64<\/p>\n

This isn't different for variables. For\u00a0x<\/em>3, it is equal to\u00a0xxx<\/em>. In Algebra, putting two or more variables next to each other means multiplying them.\u00a0x<\/em>\u00a0multiplied by\u00a0x<\/em>\u00a0multiplied by\u00a0x<\/em>.<\/p>\n

x<\/em>3 =\u00a0xxx<\/em><\/p>\n

Once you understand these basic\u00a0concepts, we can move on to the main topic.<\/p>\n

Multiplying X with Different Exponents<\/h2>\n

Product Rule<\/h3>\n

Multiplying X with different exponents means that you multiply the same variables\u2014in this case, \"X\"\u2014but a different amount of times.<\/p>\n

(X4) (X7) = (XXXX)(XXXXXXX)<\/p>\n

You can see that we expand the variables with exponents into different amounts of variable iterations. The number of variables written equals the value of each exponent.<\/p>\n

To simplify things further, let's remove the unnecessary parentheses.<\/p>\n

(X4) (X7) = XXXXXXXXXXX<\/p>\n

Well, that's a lot of Xs. How many X on the right side of the equal symbol? Bingo! It's 11. There are 11 copies of the same variable. You can also simplify things by replacing the result with \"X11\". Let's put all of them together.<\/p>\n

(X4) (X7) = (XXXX) (XXXXXXX)<\/p>\n

= XXXXXXXXXXX<\/p>\n

= X11<\/p>\n

Therefore, (X4)(X7) equals X11. To make this more intuitive, we can also write it as (X4)(X7) equals X(4 + 7).<\/p>\n

The calculation above introduces us to a basic exponent rule called \"Product Rule\"<\/strong>:<\/p>\n

(Xa) (Xb) = X(a + b)<\/p>\n

Keep in mind that the rule above only applies if the bases of the two exponents are the same. In the example before, the exponents of 4 and 7 have the same base: \"X\".<\/p>\n

More Examples of\u00a0Product Rule<\/h4>\n

To improve our understanding, let's pay attention to some more advanced examples of multiplying variables with different exponents.<\/p>\n

(x5 y6) (x2 y) =?<\/p>\n

To solve this problem, first, let's group the same variables.<\/p>\n

(x5 y6) (x2 y) = (x5 x2) (y6 y)<\/p>\n

Once they have grouped nicely, we can apply the product rule by adding up the exponents with the same base.<\/p>\n

(x5 x2) (y6 y) = (x(5 + 2)) (y(6 + 1))<\/p>\n

Did you see what happened? For the base of \"y\", we add\u00a0the exponent of 6 and 1, even though the second \"y\" doesn't seem to have an exponent.<\/p>\n

For any number or variable that doesn't have an exponent written above it,\u00a0it has an exponent of 1. In that case, \"y\" has an exponent of 1. The same goes for any number out there. All numbers have the power of 1 if there is no\u00a0exponent explicitly written above it.<\/p>\n

Why? Because anything to the power of 1 has 'one copy' of itself. 51 = 5; 271 = 27; y1 = y.<\/p>\n

Let\u2019s continue on with our work.<\/p>\n

(x(5 + 2)) (y(6 + 1)) = (x7) (y7)<\/p>\n

Putting all of them together, we can determine that the value of (x5 y6)(x2 y) equals to (x7)(y7).<\/p>\n

(x5 y6) (x2 y) = (x5 x2) (y6 y)<\/p>\n

= (x(5 + 2)) (y(6 + 1))<\/p>\n

= (x7) (y7)<\/p>\n

Quotient Rule<\/h3>\n

If the Product Rule lets us add two exponents together, the Quotient Rule would instead grant us the ability subtract two exponents. Consider this expression.<\/p>\n

X6 \/ X3 = ?<\/p>\n

Can you determine the result?<\/p>\n

By this point, you might have understood that we can expand the bases to reveal their copies based on the value of each exponent.<\/p>\n

X6 \/ X3 = XXXXXX \/ XXX<\/p>\n

From here, we can erase three Xs from both the numerator and denominator of the fraction.<\/p>\n

We can gather all of the steps together to see the full picture.<\/p>\n

X6 \/ X3 = XXXXXX \/ XXX<\/p>\n

= XXX<\/p>\n

= X3<\/p>\n

Thus, X6\/X3 equals X3. Or, we can also write it as X6\/X3 equals X(6 - 3).<\/p>\n

Once again, our work brings us to the second rule of exponents: the\u00a0Quotient Rule<\/strong>.<\/p>\n

Xa \/ Xb = X(a \u2013 b)<\/p>\n

Identical to the Product Rule, the Quotient Rule can only work if all of the bases have the same value. In our case, the bases are the same variable of \"X\".<\/p>\n

Power Rule<\/h3>\n

Our next topic looks at a base with a visible exponent raised to a certain power. Take a look at the equation below.<\/p>\n

(X3)4 =?<\/p>\n

Can you determine how many Xs that is?<\/p>\n

If it's too confusing, let's break it down into a more tangible form.<\/p>\n

(X3)4 = (XXX)4<\/p>\n

First, we expand the exponent inside the parentheses. This way, we can assign the outer exponent of \"4\" into each X.<\/p>\n

(XXX)4 = (X4) (X4) (X4)<\/p>\n

Finally, using the Product Rule, let's combine the exponents into one.<\/p>\n

(X3)4 = (XXX)4<\/p>\n

= (X4) (X4) (X4)<\/p>\n

= X(4 + 4 + 4)<\/p>\n

= X12<\/p>\n

Thus, we can determine that (X3)4 equals X12. Writing it another way, we can see that (X3)4 equals to X(3 x 4)<\/p>\n

Thanks to the calculation above, we can familiarize ourselves with the third essential rule of exponents: the\u00a0Power Rule<\/strong>.<\/p>\n

(Xa)b = X(ab)<\/p>\n

More Examples with Power Rule<\/h4>\n

Let's have a look at a more complicated example.<\/p>\n

Can you simplify this?<\/p>\n

(x5y9z2)3<\/p>\n

First, remember that all bases have different variables so we can't add exponents together using the Product Rule.<\/p>\n

In that case, using the Power Rule, we can instead multiply the inner exponents with the outer exponent. With this method, we can easily and quickly see the result.<\/p>\n

(x5 y9 z2)3 = (x(5 x 3)) (y(9 x 3)) (z(2 x 3))<\/p>\n

= x15 y27 z6<\/p>\n

Zero Rule<\/h3>\n

Before we dive into the Zero rule, consider the expression below.<\/p>\n

X4 \/ X4 = ?<\/p>\n

At a single glance, we can see that both the numerator and denominator are the same value. The division of anything with the same value equals 1. Thus, the result of the equation is 1.<\/p>\n

X4 \/ X4 = 1<\/p>\n

With that said, do you remember the Quotient Rule? The rule states that we can subtract two exponents if two powers with the same bases are divided.<\/p>\n

X4 \/ X4 = X(4 -4)<\/p>\n

= X0<\/p>\n

This calculation\u00a0brings us to the\u00a0Zero Rule<\/strong>.<\/p>\n

X0 = 1<\/p>\n

X \u2260 0<\/p>\n

The condition of X \u2260 0 is there since 0 divided by 0 is undefined. Additionally, our previous calculation is only valid if X is not 0.<\/p>\n

Negative Exponents<\/h3>\n

Even exponents may come in a negative form. Despite how it sounds, however, it's not a super complicated concept. It\u00a0only means that the base should be on the opposite side of the fraction line. Take a look at the example below.<\/p>\n

X-3 = 1 \/ X3<\/p>\n

As you can see, we flip the base from the numerator position to the denominator or vice versa to turn a negative exponent into a positive one.<\/p>\n

With this knowledge, can you solve this equation?<\/p>\n

2x-3 \/ x-5 = ?<\/p>\n

Take it slow and keep in mind that we only need to reverse the position of the base.<\/p>\n

2x-3 \/ x-5 = 2x5 \/ x3<\/p>\n

We do not flip the number \"2\" because it has the power of 1.<\/p>\n

Final Words<\/h2>\n

When working with exponent problems, you might have different sequences or steps compared to other people. This is fine. Math is pretty flexible, so you don't have to force yourself to use the same steps to solve problems, as long as your answer is correct.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

<\/div>\n

Be the first to comment<\/a> below.<\/p>\n","protected":false},"excerpt":{"rendered":"

Do you still remember the concept of variables and exponents? Before we jump into the topic in question, let's review. x3 x\u00a0is a\u00a0variable,\u00a0or something that has an unknown value. The value of\u00a0x\u00a0may vary\u00a0(hence the name). In one equation,\u00a0x\u00a0may be equal to 4. In another equation,\u00a0x\u00a0may be equal to 7 or a more complicated number like […]<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mo_disable_npp":""},"categories":[4],"tags":[],"_links":{"self":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/12468"}],"collection":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/comments?post=12468"}],"version-history":[{"count":1,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/12468\/revisions"}],"predecessor-version":[{"id":12469,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/posts\/12468\/revisions\/12469"}],"wp:attachment":[{"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/media?parent=12468"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/categories?post=12468"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.intmath.com\/blog\/wp-json\/wp\/v2\/tags?post=12468"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}