# LAWS OF EXPONENTS

Law 1  :

The product of two powers with the same base equals that base raised to the sum of the exponents.

If x is any nonzero real number and m and n are integers, then

xm ⋅ xn  =  xm+n

Example :

34 ⋅ 35  =  34+5

=  39

Law 2 :

A power raised to another power equals that base raised to the product of the exponents.

If x is any nonzero real number and m and n are integers, then

(xm)n  =  xmn

Example :

(32)4  =  3(2)(4)

=  38

Law 3 :

A product raised to a power equals the product of each factor raised to that power.

If x and y are any nonzero real numbers and m is any integer, then

(xy)m  =  xm ⋅ ym

Example :

(3 ⋅ 5)2  =  32 ⋅ 52

=  9 ⋅ 25

=  225

Law 4 :

The quotient of two non zero powers with the same base equals the base raised to the difference of the exponents.

If x is any nonzero real number and m and n are integers, then

xm ÷ xn  =  xm-n

Example :

37 ÷ 35  =  37-5

=  32

=  9

Law 5 :

A quotient raised to a positive power equals the quotient of each base raised to that power.

If x and y are any nonzero real numbers and m is a positive integer, then

(x/y)m  =  xm/ym

Example :

(3/5)2  =  32/52

=  9/25

Law 6 :

A quotient raised to a negative power equals the reciprocal of the quotient raised to the opposite (positive) power.

If x and y are any nonzero real numbers and m is a positive integer, then

(x/y)-m  =  (y/x)m

Example :

(5/3)-2  =  (3/5)2

=  32/52

=  9/25

Law 7 :

If a power is moved from numerator to denominator or denominator to numerator, the sign of the exponent has to be changed.

That is

x-m  =  1/xm

Example :

3-2  =  1/32

3-2  =  1/9

Law 8 :

For any nonzero base, if the exponent is zero, its value is 1.

x0  =  1

Example :

30  =  1

Law 9 :

For any base base, if there is no exponent, the exponent is assumed to be 1.

x1  =  x

Example :

31  =  3

Law 10 :

If an exponent is transferred from one side of the equation to the other side of the equation, reciprocal of the exponent has to be taken.

xm/n  =  y -----> x  =  yn/m

Example :

x1/2  =  3

x  =  32/1

x  =  32

x  =  9

Law 11 :

If two powers are equal with the same base, exponents can be equated.

ax  =  ay -----> x  =  y

Example :

3m  =  35 -----> m  =  5

Law 12 :

If two powers are equal with the same exponent, bases can be equated.

xa  =  ya -----> x  =  y

Example :

k3  =  53 -----> k  =  5

## Important Note

Many students do not know the difference between

(-3)2   and   -32

Order of operations (PEMDAS) dictates that parentheses take precedence.

So, we have

(-3)=  (-3) ⋅ (-3)

(-3)2  =  9

Without parentheses, exponents take precedence :

-32  =  -3 ⋅ 3

-32  =  -9

The negative is not applied until the exponent operation is carried through. We have to make sure that we understand this. So, we will not make this common mistake.

Sometimes, the result turns out to be the same, as in.

(-2)3   and   -23

We have to make sure why they yield the same result.

## Solved Problems

Problem 1 :

If a-1/2 = 5, then find the value of a.

Solution :

a-1/2  =  5

a  =  5-2/1

a  =  5-2

a  =  1/52

a  =  1/25

Problem 2 :

If n = 12 + 14 + 16 + 18 +...............+ 150, then find the value of n.

Solution :

1 to the power of anything is equal to 1.

So, we have

1=  1

14  =  1

16  =  1

and so on.

List out all the exponents.

2, 4, 6, 8, ................ 48, 50

Divide each element by 2.

1, 2, 3, 4, ................ 24, 25

We can clearly see there are 25 terms in the series shown below.

12 + 14 + 16 + 18 +...............+ 150

Therefore, n is the sum of twenty-five 1's.

That is,

n  =  12 + 14 + 16 + 18 +...............+ 150

n  =  25

Problem 3 :

If 42n + 3 = 8n + 5, then find the value of n.

Solution :

42n + 3  =  8n + 5

(22)2n + 3  =  (23)n + 5

22(2n + 3)  =  23(n + 5)

Equate the exponents.

2(2n + 3)  =  3(n + 5)

4n + 6  =  3n + 15

n  =  9

Problem 4 :

If 2x/2= 23, then find the value x in terms of y.

Solution :

2x / 2y  =  23

2x - y  =  23

x - y  =  3

x  =  y + 3

Problem 5 :

If ax = b, by = c and  cz = a, then find the value of xyz.

Solution :

Let

ax  =  b -----(1)

by  =  c -----(2)

cz  =  a -----(3)

Substitute a  =  cz in (1).

(1)-----> (cz)x  =  b

czx  =  b

Substitute c  =  by.

(by)zx  =  b

bxyz  =  b

bxyz  =  b1

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