When we are presented with exercises or expressions where multiplication of powers with the same base appears, we can add the exponents.

The result obtained from adding the exponents will be the new exponent and the original base is maintained.

##### The formula of the rule:$a^m\times a^n=a^{(m+n)}$

It doesn't matter how many terms there are. As long as there are products of powers with the same base, we can add their exponents and obtain a new one that we apply to the base.

It is important to remember that this property should only be applied when there are products of powers with the same base. In other words, if we have a multiplication of powers with different bases, we cannot add the exponents.

This property also pertains to algebraic expressions.

## Examples with solutions for Multiplication of Powers

### Exercise #1

$4^2\times4^4=$

### Step-by-Step Solution

To solve the exercise we use the property of multiplication of powers with the same bases:

$a^n * a^m = a^{n+m}$

With the help of this property, we can add the exponents.

$4^2\times4^4=4^{4+2}=4^6$

$4^6$

### Exercise #2

$7^9\times7=$

### Step-by-Step Solution

According to the property of powers, when there are two powers with the same base multiplied together, the exponents should be added.

According to the formula:$a^n\times a^m=a^{n+m}$

It is important to remember that a number without a power is equivalent to a number raised to 1, not to 0.

Therefore, if we add the exponents:

$7^{9+1}=7^{10}$

$7^{10}$

### Exercise #3

$2^{10}\cdot2^7\cdot2^6=$

### Step-by-Step Solution

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Keep in mind that this property is also valid for several terms in the multiplication and not just for two, for example for the multiplication of three terms with the same base we obtain:

$a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}$When we use the mentioned power property twice, we could also perform the same calculation for four terms of the multiplication of five, etc.,

Keep in mind that all the terms in the multiplication have the same base, so we will use the previous property:

$2^{10}\cdot2^7\cdot2^6=2^{10+7+6}=2^{23}$Therefore, the correct answer is option c.

$2^{23}$

### Exercise #4

$8^2\cdot8^3\cdot8^5=$

### Step-by-Step Solution

All bases are equal and therefore the exponents can be added together.

$8^2\cdot8^3\cdot8^5=8^{10}$

$8^{10}$

### Exercise #5

$5^4\times25=$

### Step-by-Step Solution

To solve this exercise, first we note that 25 is the result of a power and we reduce it to a common base of 5.

$\sqrt{25}=5$$25=5^2$Now, we go back to the initial exercise and solve by adding the powers according to the formula:

$a^n\times a^m=a^{n+m}$

$5^4\times25=5^4\times5^2=5^{4+2}=5^6$

$5^6$

### Exercise #6

$a^3\times a^4=$

### Step-by-Step Solution

Here, we will need to calculate a multiplication between terms with identical bases, therefore we will use the appropriate power property:

$b^m\cdot b^n=b^{m+n}$Note that this property can only be used to calculate the multiplication between terms with identical bases,

We apply it to the problem:

$a^3\cdot a^4=a^{3+4}=a^7$Therefore, the correct answer is option b.

$a^7$

### Exercise #7

$x^2\times x^5=$

### Step-by-Step Solution

Here we will have to to multiply terms with identical bases, therefore we use the appropriate power property:

$b^m\cdot b^n=b^{m+n}$Note that this property can only be used to calculate the multiplication between terms with identical bases,

From now on we no longer write the multiplication sign, but use the accepted form of writing in which placing terms next to each other means multiplication.

We apply it in the problem:

$x^2x^5=x^{2+5}=x^7$Therefore, the correct answer is D.

$x^7$

### Exercise #8

$2^{2x+1}\cdot2^5\cdot2^{3x}=$

### Step-by-Step Solution

Since the bases are the same, the exponents can be added:

$2x+1+5+3x=5x+6$

$2^{5x+6}$

### Exercise #9

$4^{2y}\cdot4^{-5}\cdot4^{-y}\cdot4^6=$

### Step-by-Step Solution

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply the property for this problem:

$4^{2y}\cdot4^{-5}\cdot4^{-y}\cdot4^6= 4^{2y+(-5)+(-y)+6}=4^{2y-5-y+6}$We simplify the expression we got in the last step:

$4^{2y-5-y+6} =4^{y+1}$When we add similar terms in the exponent.

Therefore, the correct answer is option c.

$4^{y+1}$

### Exercise #10

$7^5\cdot7^{-6}=\text{?}$

### Step-by-Step Solution

We begin by using the rule for multiplying exponents. (the multiplication between terms with identical bases):

$a^m\cdot a^n=a^{m+n}$We then apply it to the problem:

$7^5\cdot7^{-6}=7^{5+(-6)}=7^{5-6}=7^{-1}$When in a first stage we begin by applying the aforementioned rule and then continue on to simplify the expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression obtained in the previous step:

$7^{-1}=\frac{1}{7^1}=\frac{1}{7}$We then summarise the solution to the problem: $7^5\cdot7^{-6}=7^{-1}=\frac{1}{7}$Therefore, the correct answer is option B.

$\frac{1}{7}$

### Exercise #11

$7^{2x+1}\cdot7^{-1}\cdot7^x=$

### Step-by-Step Solution

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply the property to our expression:

$7^{2x+1}\cdot7^{-1}\cdot7^x=7^{2x+1+(-1)+x}=7^{2x+1-1+x}$We simplify the expression we got in the last step:

$7^{2x+1-1+x}=7^{3x}$When we add similar terms in the exponent.

Therefore, the correct answer is option d.

$7^{3x}$

### Exercise #12

$12^4\cdot12^{-6}=\text{?}$

### Step-by-Step Solution

We begin by using the power rule of exponents; for the multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply it to the given problem:

$12^4\cdot12^{-6}=12^{4+(-6)}=12^{4-6}=12^{-2}$When in a first stage we apply the aforementioned rule and then simplify the subsequent expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression that we obtained in the previous step:

$12^{-2}=\frac{1}{12^2}=\frac{1}{144}$Lastly we summarise the solution to the problem: $12^4\cdot12^{-6}=12^{-2} =\frac{1}{144}$Therefore, the correct answer is option A.

$\frac{1}{144}$

### Exercise #13

Solve the exercise:

$Y^2+Y^6-Y^5\cdot Y=$

### Step-by-Step Solution

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply it in the problem:

$Y^2+Y^6-Y^5\cdot Y=Y^2+Y^6-Y^{5+1}=Y^2+Y^6-Y^6=Y^2$When we apply the previous property to the third expression from the left in the sum, and then simplify the total expression by adding like terms.

Therefore, the correct answer is option D.

$Y^2$

### Exercise #14

$x^3\cdot x^4\cdot\frac{2}{x^3}\cdot x^{-8}=\text{?}$

### Step-by-Step Solution

First we will rearrange the expression and use the fact that multiplying a fraction means multiplying the numerator of the fraction, and the distributive property of multiplication:

$x^3\cdot x^4\cdot\frac{2}{x^3}\cdot x^{-8}=x^3\cdot x^4\cdot 2\cdot \frac{1}{x^3}\cdot x^{-8}=2\cdot x^3\cdot x^4\cdot \frac{1}{x^3}\cdot x^{-8}$Next, we'll use the law of exponents for negative exponents:

$a^{-n}=\frac{1}{a^n}$We'll apply the law of exponents to the expression in the problem:

$2\cdot x^3\cdot x^4\cdot\frac{1}{x^3}\cdot x^{-8} =2\cdot x^3\cdot x^4\cdot x^{-3}\cdot x^{-8}$When we applied the above law of exponents for the fraction in the multiplication,

From now on, we will no longer use the multiplication sign and will switch to the conventional notation where juxtaposition of terms means multiplication between them,

Now we'll recall the law of exponents for multiplying terms with the same base:

$a^m\cdot a^n=a^{m+n}$And we'll apply this law of exponents to the expression we got in the last step:

$2x^3x^4x^{-3}x^{-8} =2x^{3+4+(-3)+(-8)} =2x^{3+4-3-8}=2x^{-4}$When in the first stage we applied the above law of exponents and in the following stages we simplified the expression in the exponent,

Let's summarize the solution steps so far, we got that:
$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}= 2 x^3x^4 x^{-3} x^{-8} =2x^{-4}$

Now let's note that there is no such answer in the given options, a further check of what we've done so far will also reveal that there is no calculation error,

Therefore, we can conclude that additional mathematical manipulation is required to determine which is the correct answer among the suggested answers,

Let's note that in answers A and B there are similar expressions to the one we got in the last stage, however - we can directly rule out the other two options since they are clearly different from the expression we got,

Furthermore, we'll note that in the expression we got, x is in a negative exponent and is in the numerator (Note at the end of the solution on this topic), whereas in answer B it is in a positive exponent and in the denominator (and both are in the numerator - Note at the end of the solution on this topic), so we'll rule out this answer,

If so - we are left with only one option - which is answer A', however we want to verify (and need to verify!) that this is indeed the correct answer:

Let's note that in the expression we got x is in a negative exponent and is in the numerator (Note at the end of the solution on this topic), whereas in answer B it is in a positive exponent and in the denominator , which reminds us of the law of exponents for negative exponents mentioned at the beginning of the solution,

In addition, let's note that in answer B x is in the second power but inside parentheses that are also in the second power, whereas in the expression we got in the last stage of solving the problem x is in the fourth power which might remind us of the law of exponents for power to a power,

We'll check this, starting with the law of exponents for negative exponents mentioned at the beginning of the solution, but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$Next, we'll represent the term with the negative exponent that we got in the last stage of solving the problem, as a term in the denominator of the fraction with a positive exponent:

$2x^{-4}=2\cdot\frac{1}{x^4}$When we applied the above law of exponents,

Next, let's note that using the law of exponents for power to a power, but in the opposite direction:

$a^{m\cdot n}= (a^m)^n$We can conclude that:

$x^4=x^{2\cdot2}=(x^2)^2$Therefore, we'll return to the expression we got in the last stage and apply this understanding:

$2\cdot\frac{1}{x^4} =2\cdot\frac{1}{(x^2)^2}$ Let's summarize then the problem-solving stages so far, we got that:

$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4} =2\cdot\frac{1}{(x^2)^2}$Let's note that we still haven't got the exact expression suggested in answer A, but we are already very close,

To reach the exact expression claimed in answer A, we'll recall another important law of exponents, and a useful mathematical fact:

Let's recall the law of exponents for exponents applying to terms in parentheses, but in the opposite direction:

$\frac{a^n}{c^n}=\big(\frac{a}{c}\big)^n$And let's also recall the fact that raising the number 1 to any power will yield the result 1:

$1^{x}=1$And therefore we can write the expression we got in the last stage in the following way:

$2\cdot\frac{1}{(x^2)^2}=2\cdot\frac{1^2}{(x^2)^2}$And then since in the numerator and denominator of the fraction there are terms with the same exponent we can apply the above law of exponents, and represent the fraction whose numerator and denominator are terms with the same exponent as a fraction whose numerator and denominator are the bases of the terms and it is raised to the same exponent:

$2\cdot\frac{1^2}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$ Let's summarize then the solution stages so far, we got that:

$x^3 x^4\cdot\frac{2}{x^3}\cdot x^{-8}=2x^{-4}=2\cdot\frac{1}{(x^2)^2}=2\cdot\big(\frac{1}{x^2}\big)^2$And therefore the correct answer is indeed answer A.

Note:

When it's written "the number in the numerator" despite the fact that there is no fraction in the expression at all, it's because we can always refer to any number as a number in the numerator of a fraction if we remember that any number divided by 1 equals itself, that is, we can always write a number as a fraction by writing it like this:

$X=\frac{X}{1}$And therefore we can actually refer to $X$as a number in the numerator of a fraction.

$2(\frac{1}{x^2})^2$

### Exercise #15

$\frac{9\cdot3}{8^0}=\text{?}$

### Step-by-Step Solution

We use the formula:

$a^0=1$

$\frac{9\times3}{8^0}=\frac{9\times3}{1}=9\times3$

We know that:

$9=3^2$

Therefore, we obtain:

$3^2\times3=3^2\times3^1$

We use the formula:

$a^m\times a^n=a^{m+n}$

$3^2\times3^1=3^{2+1}=3^3$

$3^3$