# 3 ways to multiply powers

In algebra, you will often come across expressions with exponents (also called powers). The power of a value is the result of multiplying that value by itself as many times as the exponent indicates. There are rules about powers, like the one where the product of two powers of the same number is a power of that number, the latter being the sum of the preceding exponents. When the bases are different, the resolution of the problem comes down to a direct calculation.

## Steps

### Method 1 of 3: Multiply powers with the same base

#### Step 1. Make sure the bases are the same

The base is a value, literal or numeric, raised to a power and placed on the main writing line. The method detailed here only works with expressions having the same base.

• You will be able to use it with 52 × 53 { displaystyle 5 ^ {2} times 5 ^ {3}}

, car les deux bases sont identiques, à savoir 5. Par contre, simplifier 52×23{displaystyle 5^{2}\times 2^{3}}

avec cette méthode est impossible, car les bases sont différentes (5 et 2).

#### Step 2. Add the exponents

To do this, you will keep the base to which you will assign the sum of the exponents as the exponent.

• Let's take the example 52 × 53 { displaystyle 5 ^ {2} times 5 ^ {3}}

. Vous allez garder la base (5) et additionnez les exposants, ce qui donne:

52×53{displaystyle 5^{2}\times 5^{3}}

52×53=52+3{displaystyle 5^{2}\times 5^{3}=5^{2+3}}

52×53=55{displaystyle 5^{2}\times 5^{3}=5^{5}}

#### Step 3. Calculate the expression found

The exponent tells you how many times you should multiply the base by itself. Depending on the case, the power calculation will be done head-on or by hand, but if the base is a large number, you will use a calculator.

• In our example, 55 = 5 × 5 × 5 × 5 × 5 { displaystyle 5 ^ {5} = 5 \ times 5 \ times 5 \ times 5 \ times 5}

donc 55=3 125{displaystyle 5^{5}=3\ 125}

En résumé, 52×53=3 125{displaystyle 5^{2}\times 5^{3}=3\ 125}

### Méthode 2 sur 3: Multiplier des puissances ayant des bases différentes

#### Step 1. Calculate the first power

In theory, there is no possibility of grouping values ​​raised to a power with different bases. Depending on the case, do the calculation with a calculator or by hand. Raising a value to a power consists of multiplying that value by itself as many times as the exponent indicates.

• As an example, let's try to simplify 23 × 45 { displaystyle 2 ^ {3} times 4 ^ {5}}

. Vous avez deux valeurs différentes élevées à des puissances différentes. Calculez en premier 23{displaystyle 2^{3}}

, ce qui donne: 23=2×2×2=8{displaystyle 2^{3}=2\times 2\times 2=8}

#### Step 2. Do the same with the second expression

Multiply the base (here, 4) by itself as many times as the exponent says.

• Calculate 45 { displaystyle 4 ^ {5}}

, ce qui donne 45=4×4×4×4×4=1 024{displaystyle 4^{5}=4\times 4\times 4\times 4\times 4=1\ 024}

#### Step 3. Rewrite the now simplified operation

If we take the previous example, the product is in the form: 8 × 1 024 { displaystyle 8 \ times 1 \ 024}

#### Step 4. Do the multiplication by hand or with a calculator

You will have the answer to the problem posed.

• In our example: 8 × 1024 = 8192. { Displaystyle 8 \ times 1024 = 8 \ 192.}

Votre réponse est donc: 23×45=8 192{displaystyle 2^{3}\times 4^{5}=8\ 192}

### Méthode 3 sur 3: Multiplier les puissances de plusieurs inconnues

#### Step 1. First multiply the coefficients between them

Simply multiply them and put the result at the beginning of the expression, outside the parentheses.

• By multiplying the coefficients (2 and 8) and taking the result out of the parentheses, the expression (2x3y5) (8xy4) { displaystyle (2x ^ {3} y ^ {5}) (8xy ^ {4})}

devient:

((2)x3y5)((8)xy4)=16(x3y5)(xy4){displaystyle ((2)x^{3}y^{5})((8)xy^{4})=16(x^{3}y^{5})(xy^{4})}

#### Step 2. Add the exponents of the first variable

By taking only the powers having the same base, you add their exponents. Any variable without an exponent is assumed to have an exponent of 1.

• In our example, we group the x { displaystyle x}

:

16(x3y5)(xy4)=16(x3)y5(x)y4=16(x3+1)y5y4=16(x4)y5y4{displaystyle 16(x^{3}y^{5})(xy^{4})=16(x^{3})y^{5}(x)y^{4}=16(x^{3+1})y^{5}y^{4}=16(x^{4})y^{5}y^{4}}

#### Step 3. Add the exponents of the other variables

As before, you can only add the exponents if they have the same base and if one of the variables has no exponent, it is assumed to have one equal to 1.

• In our example, we group the y { displaystyle y}

:

16(x4)(y5y4)=16x4(y5+4)=16x4y9{displaystyle 16(x^{4})(y^{5}y^{4})=16x^{4}(y^{5+4})=16x^{4}y^{9}}

## conseils

• tout valeur, numérique ou littérale, élevée à la puissance 0 est égale à 1, ce qui fait, par exemple, que: (50)(x0)=(1)(1)=1{displaystyle (5^{0})(x^{0})=(1)(1)=1}