HEGP102 — Power Play
Complete Questions and
Answers
This guide answers the
exercises, “Figure it Out” questions, worked activities, and end-of-section
problems in the attached Grade 8 chapter. For estimation questions, the
assumptions are stated because reasonable assumptions can produce slightly
different answers.
1. Exponential growth and
paper folding
Question 1. What would the
thickness of a sheet of paper be after 30 folds?
The initial thickness is
(0.001) cm and the thickness doubles after every fold. Therefore,
[ T_n=0.001\times2^n\text{
cm}. ]
For 30 folds,
[T_{30}=0.001\times2^{30}=1,073,741.824\text{
cm}
\approx10.737\text{ km}.]
Thus, the thickness after
30 folds is approximately 10.7 km.
Question 2. What would the
thickness be after 45 and 46 folds?
[T_{45}=0.001\times2^{45}=35,184,372,088.832\text{
cm}
\approx351,843.7\text{ km},]
and
[T_{46}=0.001\times2^{46}=70,368,744,177.664\text{
cm}
\approx703,687.4\text{ km}.]
So, under the chapter’s
stated initial thickness, the thickness is about 351,844
km after 45 folds and 703,687 km after 46 folds.
The important idea is the exponential doubling, not the physical feasibility of
actually folding the paper that many times.
Question 3. Complete the
paper-thickness table.
|
Fold |
Thickness |
Fold |
Thickness |
|
18 |
262.144 cm |
24 |
167.772 m |
|
19 |
524.288 cm |
25 |
335.544 m |
|
20 |
10.486 m |
26 |
671.089 m |
|
21 |
20.972 m |
27 |
1.342 km |
|
22 |
41.943 m |
28 |
2.684 km |
|
23 |
83.886 m |
29 |
5.369 km |
|
30 |
10.737 km |
|
|
For the later table:
|
Fold |
Thickness |
Fold |
Thickness |
|
31 |
21.475 km |
36 |
687.195 km |
|
32 |
42.950 km |
37 |
1,374.390 km |
|
33 |
85.899 km |
38 |
2,748.779 km |
|
34 |
171.799 km |
39 |
5,497.558 km |
|
35 |
343.598 km |
40 |
10,995.116 km |
|
41 |
21,990.233 km |
42 |
43,980.465 km |
|
43 |
87,960.930 km |
44 |
175,921.860 km |
|
45 |
351,843.721 km |
|
|
Question 4. By how much does
the thickness increase after two folds? What happens after three folds and ten
folds?
After two folds, the
thickness is multiplied by (2^2=4). After three folds, it is multiplied by
(2^3=8). After ten folds, it is multiplied by (2^{10}=1024). Thus, the growth
is multiplicative or exponential, rather than
additive.
2. Exponential notation and
powers
Question 5. Which expression
gives the thickness after 10 folds if the initial thickness is represented by
(v)?
Answer: The correct expression is (v)
(2^{10}v), because each fold doubles the thickness.
Question 6. Express repeated
multiplication in exponential form.
[
6\times6\times6\times6=6^4, ] [ y\times y=y^2, ] [ b\times b\times b\times
b=b^4, ] [ 5\times5\times7\times7\times7=5^2\times7^3, ] [ 2\times2\times
a\times a=2^2a^2, ] [ a\times a\times a\times c\times c\times c\times c\times
d=a^3c^4d. ]
Question 7. Express 32400 as
a product of prime powers.
[
32400=2^4\times3^4\times5^2. ]
Question 8. Find the values
of the following.
|
Expression |
Value |
|
(2\times10^3) |
2000 |
|
(7^2\times2^3) |
392 |
|
(3\times4^4) |
768 |
|
((-3)^2\times(-5)^2) |
225 |
|
(3^2\times10^4) |
90,000 |
|
((-2)^5\times(-10)^6) |
32,000,000 |
The last result is positive
because both factors are negative and their product is positive.
Question 9. What are
((-1)^5), ((-1)^{56}), and ((-2)^4)? What are (0^2), (0^5), and (0^n)?
[
(-1)^5=-1\quad\text{(negative)}, ] [ (-1)^{56}=1\quad\text{(positive)}, ] [
(-2)^4=16, ] [ 0^2=0,\quad0^5=0,\quad0^n=0\text{ for every positive integer }n.
]
Question 10. How many rooms
and diamonds are in “The Stones that Shine”?
There are three daughters,
each daughter has three baskets, each basket has three keys, and each key opens
three rooms. Therefore,
[ \text{rooms}=3^4=81. ]
Each room has three tables,
each table has three necklaces, and each necklace has three diamonds. Hence,
[ \text{diamonds}=3^7=2187.
]
Question 11. Why can (3^7)
be written as (3^2\times3^5)?
Because the seven factors
of 3 can be split into groups of two and five:
[ 3^7=(3\times3)(3\times3\times3\times3\times3)=3^2\times3^5.
]
This illustrates the rule
[ n^a\times n^b=n^{a+b}. ]
Question 12. Use the
exponent rule to calculate (2^9), (5^7), and (4^6).
[
2^9=2^4\times2^5=16\times32=512, ] [ 5^7=5^3\times5^4=125\times625=78,125, ] [
4^6=4^3\times4^3=64\times64=4096. ]
Question 13. Write each
expression as a power of a power in at least two ways.
[
8^6=(8^2)^3=(8^3)^2=(2^3)^6=2^{18}, ] [ 7^{15}=(7^3)^5=(7^5)^3, ] [
9^{14}=(9^2)^7=(9^7)^2=(3^2)^{14}=3^{28}, ] [ 5^8=(5^2)^4=(5^4)^2. ]
The rule used is
[ (n^a)^b=n^{ab}. ]
3. Magical Pond and
combinations
Question 14. When was the
doubling pond half full?
If the pond was completely
full on the 30th day and the number of lotuses doubled each day, it was half
full on the 29th day. The number of lotuses was:
[ \text{fully
covered}=2^{30},\qquad \text{half covered}=2^{29}. ]
Question 15. A lotus is
placed in a pond where flowers triple daily after receiving the flowers from a
doubling pond after four days. How many flowers are present after four more
days?
After four days in the
doubling pond:
[ 2^4=16. ]
After four days in the
tripling pond:
[
16\times3^4=2^4\times3^4=(2\times3)^4=6^4=1296. ]
There will be 1296 lotuses.
The order does not matter
because multiplication is commutative:
[
2^4\times3^4=3^4\times2^4=6^4. ]
Question 16. Find
(2^5\times5^5).
[ 2^5\times5^5=(2\times5)^5=10^5=100,000.
]
Question 17. Simplify
(10^4\div5^4).
[
\frac{10^4}{5^4}=\left(\frac{10}{5}\right)^4=2^4=16. ]
Question 18. Roxie has 7
dresses, 2 hats, and 3 pairs of shoes. How many outfits are possible?
For each dress she can
choose one of two hats and one of three pairs of shoes:
[ 7\times2\times3=42. ]
Therefore, Roxie can dress
in 42 different ways.
Question 19. How many
five-digit passwords are possible when digits 0–9 may be repeated?
Each of the five positions
has 10 choices:
[ 10^5=100,000. ]
Therefore, there are 100,000 passwords.
Question 20. How many
passwords are possible with a six-slot lock using the letters A–Z?
Each slot has 26 choices:
[ 26^6=308,915,776. ]
Therefore, there are 308,915,776 possible passwords.
4. Negative exponents and
powers of 7
Question 21. What is
(2^{100}\div2^{25}) in powers of 2?
[
2^{100}\div2^{25}=2^{100-25}=2^{75}. ]
Question 22. Why is the base
not allowed to be zero in (x^0=1)?
For a non-zero base,
[ x^a\div
x^a=x^{a-a}=x^0=1. ]
If (x=0), the expression
involves division by zero, which is undefined. Therefore, the standard rule
(x^0=1) applies only when (x\ne0). The expression
(0^0) is not defined in this context.
Question 23. Write
equivalent forms.
[ 2^{-4}=\frac1{2^4}, ] [
10^{-5}=\frac1{10^5}, ] [ (-7)^{-2}=\frac1{(-7)^2}, ] [
(-5)^{-3}=\frac1{(-5)^3}, ] [ 10^{-100}=\frac1{10^{100}}. ]
Question 24. Simplify in
exponential form.
[ 2^{-4}\times2^7=2^3, ] [ 3^2\times3^{-5}\times3^6=3^{2-5+6}=3^3,
] [ p^3\times p^{-10}=p^{-7}, ] [
2^4\times(-4)^{-2}=2^4\times(2^2)^{-2}=2^4\times2^{-4}=1, ] [
8^p\times8^q=8^{p+q}. ]
Question 25. How many times
larger is (4^2) than (4^{-2})?
[
\frac{4^2}{4^{-2}}=4^{2-(-2)}=4^4=256. ]
Therefore, (4^2) is 256 times larger than (4^{-2}).
Question 26. Complete the
powers-of-7 calculations.
|
Expression |
Answer |
|
(2401\times49) |
(7^6) |
|
(49^3) |
(7^6) |
|
(343\times2401) |
(7^7) |
|
(16,807\div49) |
(7^3) |
|
(7\div343) |
(7^{-2}) |
|
(16,807\div8,23,543) |
(7^{-2}) |
|
(1,17,649\times\frac1{343}) |
(7^3) |
|
(\frac1{343}\times\frac1{343}) |
(7^{-6}) |
5. Powers of 10 and
scientific notation
Question 27. Write the
following in expanded form using powers of 10.
[
172=1\times10^2+7\times10^1+2\times10^0, ]
[
5642=5\times10^3+6\times10^2+4\times10^1+2\times10^0, ]
[
6374=6\times10^3+3\times10^2+7\times10^1+4\times10^0. ]
Question 28. Which is the
smallest: the Sun–Saturn, Saturn–Uranus, or Sun–Earth distance?
The distances are
approximately:
[ 1.4335\times10^{12}\text{
m},\quad1.439\times10^{12}\text{ m},\quad1.496\times10^{11}\text{ m}. ]
The smallest is the Sun–Earth distance, because (10^{11}) is one-tenth the
order of (10^{12}).
Question 29. Mark the
Earth’s position relative to the Sun and Saturn.
The Sun–Earth distance is
(1.496\times10^{11}) m, while the Sun–Saturn distance is (1.4335\times10^{12})
m. The Earth is therefore close to the Sun compared with Saturn, at roughly
10.4% of the distance from the Sun to Saturn:
[ \frac{1.496\times10^{11}}{1.4335\times10^{12}}\approx0.104.
]
The order is:
[ \textbf{Sun — Earth
————————————— Saturn}. ]
Question 30. Express the
following numbers in scientific notation.
|
Number |
Scientific notation |
|
59,853 |
(5.9853\times10^4) |
|
65,950 |
(6.595\times10^4) |
|
34,30,000 |
(3.43\times10^6) |
|
70,04,00,00,000 |
(7.004\times10^{10}) |
6. Estimation and modelling
activities
Question 31. Estimate the
worth of jaggery equal to Roxie’s weight and wheat equal to Estu’s weight.
Using the chapter’s
assumptions—Roxie weighs 45 kg and jaggery costs ₹70 per kg—
[ 45\times70=₹3150. ]
Using Estu’s assumed weight
of 50 kg and wheat cost of ₹50 per kg,
[ 50\times50=₹2500. ]
Thus, the estimated values
are ₹3,150 of jaggery and ₹2,500
of wheat.
Question 32. Approximately
how many one-rupee coins equal Roxie’s weight?
Assume Roxie weighs 45 kg
and one ₹1 coin weighs approximately 3.85 g. Then
[ 45\text{ kg}=45,000\text{
g}, ] [ \text{number of coins}\approx\frac{45,000}{3.85}\approx11,688. ]
So, approximately 11,700 one-rupee coins would equal her weight. The result
depends on the assumed coin mass.
Question 33. How much money
would the equivalent weight be in ₹5 coins or ₹10 notes?
Using the same approximate
mass of 3.85 g per ₹5 coin, about 11,700 coins would be needed. Their monetary
value would be approximately
[ 11,700\times₹5=₹58,500. ]
For ₹10 notes, assume one
note weighs about 1 g. Then approximately 45,000 notes would be needed, with
value
[
45,000\times₹10=₹4,50,000. ]
These are estimates, not
exact values.
Question 34. How many people
might benefit from notebooks or food worth Estu’s or Roxie’s weight?
One reasonable model is to
assume each notebook costs ₹50 and each meal costs ₹50. Using ₹2,500 worth of
notebooks or food:
[ ₹2500\div₹50=50. ]
Therefore, approximately 50 people could receive one notebook or one meal under
these assumptions.
Question 35. If a 400 km
pilgrimage is walked at 5 km/h for 8 hours per day, how long does it take?
The walking time is
[ 400\div5=80\text{ hours}.
]
At 8 hours per day,
[ 80\div8=10\text{ days}. ]
Thus, the journey would
take about 10 walking days, excluding rest days.
Question 36. How many times
could a person walk around Earth in a lifetime?
Assume Earth’s
circumference is 40,000 km, a person walks 5 km/h for 8 hours daily, and the
person walks for 70 years.
Daily distance:
[ 5\times8=40\text{ km}. ]
Total distance in 70 years:
[
40\times365.25\times70\approx1,022,700\text{ km}. ]
Number of
circumnavigations:
[
1,022,700\div40,000\approx25.6. ]
Therefore, the person could
walk around Earth approximately 25 times under
these assumptions.
Question 37. How many steps
would a ladder to the Moon need if each step is 20 cm?
The Earth–Moon distance is
approximately 384,400 km. Convert to centimetres:
[ 384,400\text{
km}=38,440,000,000\text{ cm}. ]
With steps 20 cm apart,
[
\frac{38,440,000,000}{20}=1,922,000,000. ]
Therefore, approximately 1,922,000,000 steps, or 192 crore 20
lakh steps, would be needed. This is linear growth because the same
fixed distance is added for every step.
Question 38. Give examples
of linear and exponential growth.
Linear growth includes
saving ₹100 every day, walking 5 km every hour, or adding 20 cm for every
ladder step. The increase is a fixed amount.
Exponential growth includes
paper thickness doubling after each fold, a population doubling each period, or
bacteria multiplying by a fixed factor. The quantity is multiplied by a fixed
factor.
7. Large numbers and time
scales
Question 39. Complete the
missing scientific-notation values.
The global starling
population is approximately 1.3 billion:
[ 1.3\text{
billion}=1.3\times10^9. ]
The global mosquito
population is approximately 110 trillion:
[ 110\text{
trillion}=1.1\times10^{14}. ]
A fossil dated to 15
million years ago corresponds to approximately
[15,000,000\text{
years}\times31,700,000\text{ seconds/year}
\approx4.76\times10^{14}\text{ seconds}.]
Question 40. How many ants
are there for every human?
Using approximately
(2\times10^{16}) ants and (8\times10^9) humans,
[
\frac{2\times10^{16}}{8\times10^9}=0.25\times10^7=2.5\times10^6. ]
There are approximately (2.5\times10^6) ants per human, or 2.5
million ants per human.
Question 41. If a flock
contains 10,000 starlings, how many flocks could there be?
Using a global population
of (1.3\times10^9) starlings,
[
\frac{1.3\times10^9}{10^4}=1.3\times10^5. ]
There could be
approximately (1.3\times10^5) flocks, or 130,000
flocks.
Question 42. If each tree
has about (10^4) leaves, how many leaves are on all trees?
Using approximately
(3\times10^{12}) trees,
[ 3\times10^{12}\times10^4=3\times10^{16}.
]
There are approximately (3\times10^{16}) leaves.
Question 43. How many sheets
of paper would reach the Moon?
Using the chapter’s paper
thickness of (0.001) cm and the Earth–Moon distance of 384,400 km:
[ 384,400\text{
km}=3.844\times10^{10}\text{ cm}, ]
[
\frac{3.844\times10^{10}}{0.001}=3.844\times10^{13}. ]
Therefore, approximately (3.844\times10^{13}) sheets would be needed under the
chapter’s stated thickness assumption.
Question 44. Roxie is 4840
days old. How many hours old is she?
[
4840\times24=116,160\text{ hours}. ]
Roxie is 116,160 hours old.
Question 45. Estu is 4070
days old. What is his approximate date of birth?
Using the current date in
this session, 17 August 2026, subtracting 4070 days
gives approximately 26 June 2015. The exact date
depends on the reference date and whether the stated age includes the current
day.
Question 46. If you have
lived for one million seconds, how old are you?
[ 1,000,000\div(60\times60\times24)\approx11.57\text{
days}. ]
Thus, one million seconds
is approximately 11.6 days, or about 12 days.
Question 47. Give examples
of events of the order of (10^5) and (10^6) seconds.
(10^5) seconds is
approximately 1.16 days. Examples include the duration of a long sporting event
or a short multi-day journey.
(10^6) seconds is
approximately 11.6 days. Examples include a school holiday, a two-week
expedition, or a multi-day scientific observation.
8. “Try This” questions on
very large quantities
Question 48. If one star is
counted every second, how long would it take to count all stars in the
observable universe?
Using approximately
(2\times10^{23}) stars and one second per star:
[
\text{time}=2\times10^{23}\text{ seconds}. ]
It would take approximately
(2.0\times10^{23}) seconds.
Question 49. If one drinks
200 ml of water every 10 seconds, how long would it take to drink all Earth’s
water?
Using approximately
(2\times10^{25}) drops and 16 drops per millilitre, the total volume is
[
\frac{2\times10^{25}}{16}=1.25\times10^{24}\text{ ml}. ]
At 200 ml every 10 seconds,
the time is
[\frac{1.25\times10^{24}}{200}\times10
=6.25\times10^{22}\text{ seconds}.]
Therefore, it would take
approximately (6.25\times10^{22}) seconds.
Question 50. What does the
first part of names such as million, billion, trillion, and quadrillion denote?
The prefixes indicate
successive powers of 1000:
[ 10^6\text{
million},\quad10^9\text{ billion},\quad10^{12}\text{
trillion},\quad10^{15}\text{ quadrillion}. ]
Each step increases the
exponent by 3 because the number is multiplied by 1000.
9. Figure it Out
Question 1. Find the units
digit of (2^{224}\div4^{32}).
Since
(4^{32}=(2^2)^{32}=2^{64}),
[
\frac{2^{224}}{4^{32}}=2^{224-64}=2^{160}. ]
The units digits of powers
of 2 repeat 2, 4, 8, 6. Since 160 is divisible by 4, the units digit is 6.
Question 2. Five bottles are
in each container, and one new container arrives each day. How many bottles are
there after 40 days?
[
5\times40=200=2\times10^2. ]
There are 200 bottles.
Question 3. Write each
number as a product of powers in three different ways.
(i) (64^3)
Since (64=2^6=4^3=8^2),
examples are:
[
64^3=2^{18}=2^{10}\times2^8, ] [ 64^3=4^5\times4^4, ] [ 64^3=8^3\times8^3. ]
(ii) (192^8)
Since (192=2^6\times3),
[ 192^8=2^{48}\times3^8. ]
Three valid forms are:
[ 2^{48}\times3^8, ] [ 2^{40}\times2^8\times3^8,
] [ 2^{40}\times6^8. ]
(iii) (32^{-5})
Since (32=2^5),
[ 32^{-5}=2^{-25}. ]
Three valid forms are:
[ 2^{-10}\times2^{-15}, ] [
2^{-5}\times2^{-20}, ] [ 4^{-12}\times2^{-1}. ]
Question 4. Classify each
statement as Always True, Only Sometimes True, or Never True.
|
Statement |
Answer |
Reason |
|
(i) Cube numbers are also
square numbers. |
Only
Sometimes True |
A number is both when it
is a sixth power: (n^6=(n^3)^2=(n^2)^3). |
|
(ii) Fourth powers are
also square numbers. |
Always
True |
(n^4=(n^2)^2). |
|
(iii) The fifth power of
a number is divisible by its cube. |
Always
True for non-zero integers |
(n^5=n^3\times n^2). |
|
(iv) The product of two
cube numbers is a cube number. |
Always
True |
(a^3b^3=(ab)^3). |
|
(v) (q^{46}) is both a
fourth and a sixth power when q is prime. |
Never
True |
The exponent 46 is
divisible by neither 4 nor 6, so the prime exponent cannot be grouped into
fours or sixes. |
Question 5. Simplify in
exponential form.
[
10^{-2}\times10^{-5}=10^{-7}, ]
[ 5^7\div5^4=5^3, ]
[ 9^{-7}\div9^4=9^{-11}, ]
[ (13^{-2})^{-3}=13^6, ]
[
m^5n^{12}(mn)^9=m^5n^{12}m^9n^9=m^{14}n^{21}. ]
Question 6. Given
(12^2=144), find the following.
[ (1.2)^2=1.44, ] [
(0.12)^2=0.0144, ] [ (0.012)^2=0.000144, ] [ 120^2=14,400. ]
Question 7. Circle the
expressions that are equal.
The expressions are
(2^4\times3^6), (6^4\times3^2), (6^{10}), (18^2\times6^2), and (6^{24}) as
printed in the chapter’s notation. The equal expressions are:
[ 2^4\times3^6, ] [
6^4\times3^2, ] [ 18^2\times6^2. ]
Each equals (2^4\times3^6).
The expressions (6^{10}) and (6^{24}) are not equal to them.
Question 8. Identify the
greater number.
[ 4^3=64<3^4=81, ]
so (3^4)
is greater.
[ 2^8=256>8^2=64, ]
so (2^8)
is greater.
[ 100^2=10,000<2^{100},
]
so (2^{100})
is greater.
Question 9. A dairy produces
8.5 billion packets and uses digits 0–9 for unique IDs. How many digits should
each code contain?
There are (8.5\times10^9)
packets. A code of length (n) has (10^n) possibilities. Since
[ 10^9<8.5\times10^9<10^{10},
]
at least 10 digits are required.
Question 10. Which numbers
are both squares and cubes?
A number that is both a
square and a cube is a sixth power:
[ n^6=(n^3)^2=(n^2)^3. ]
Examples include
[
1^6=1,\quad2^6=64,\quad3^6=729,\quad4^6=4096. ]
There are infinitely many
such numbers.
Question 11. How many
length-five alphanumeric codes are possible?
There are 26 letters and 10
digits, so each position has 36 choices. Repetition is allowed:
[ 36^5=60,466,176. ]
Therefore, there are 60,466,176 possible codes.
Question 12. The sheep and
goat populations are each approximately (10^9). What is their total?
[ 10^9+10^9=2\times10^9. ]
The correct option is (v) (2\times10^9), which is the same as option (vi),
(10^9+10^9), before simplification.
Question 13. Calculate in
scientific notation.
(i) Total clothing pieces
Using a world population of
(8.2\times10^9) and 30 pieces per person:
[
8.2\times10^9\times30=2.46\times10^{11}. ]
Answer: (2.46\times10^{11}) pieces.
(ii) Total honeybees
100 million colonies equal
(10^8) colonies. With 50,000 bees per colony:
[
10^8\times50,000=5.0\times10^{12}. ]
Answer: (5.0\times10^{12}) bees.
(iii) Total bacterial cells
in all humans
Using 38 trillion cells per
person and (8.2\times10^9) people:
[38\times10^{12}\times8.2\times10^9
=3.116\times10^{23}.]
Answer: (3.116\times10^{23}) bacterial cells.
(iv) Total time spent eating
in a lifetime
Assume a 70-year lifetime
and one hour of eating daily:
[3600\times365\times70=91,980,000
=9.198\times10^7\text{ seconds}.]
Answer: (9.198\times10^7) seconds.
Question 14. What date was
one billion seconds ago?
One billion seconds is
approximately 31.7 years. Therefore, from any stated present date, subtract
approximately 31.7 years. Using 17 August 2026 as the reference date in this
session gives approximately late December 1994.
Because the question is date-dependent, the exact calendar date changes as the
present date changes.
10. Final activity:
Tremendous in Ten!
Question 15. Which is
greater in Round 1: 10,000,000,000,000 or (999999\times999999)?
[
10,000,000,000,000=10^{13}. ]
Also,
[
999999\times999999=(10^6-1)^2<10^{12}. ]
Therefore, 10,000,000,000,000 is greater.
Question 16. Which is
greater in Round 2?
Roxie’s expression is
[
10^{1000}+10^{1000}+10^{1000}+10^{1000}=4\times10^{1000}. ]
Estu’s expression is
[
10^{100,000,000}\times9000. ]
Since (10^{100,000,000})
has vastly more powers of 10 than (10^{1000}), Estu’s number is much greater.
Question 17. What strategies
can be used under the game’s different conditions?
With only addition and no
exponents, one can write the same digit repeatedly and add many terms. With
multiplication allowed, products such as (999999\times999999) create large
numbers quickly. If exponents are allowed, a single expression such as (9^{999999})
becomes enormously large. With all operations allowed, exponentiation dominates
ordinary addition and multiplication for sufficiently large exponents.
Important formulas
|
Rule |
Formula |
|
Product of powers with
the same base |
(n^a\times n^b=n^{a+b}) |
|
Quotient of powers with
the same base |
(n^a\div n^b=n^{a-b}) |
|
Power of a power |
((n^a)^b=n^{ab}) |
|
Product of equal powers |
(m^a\times n^a=(mn)^a) |
|
Quotient of equal powers |
(n^a\div m^a=(n/m)^a) |
|
Zero exponent |
(n^0=1), for (n\ne0) |
|
Negative exponent |
(n^{-a}=1/n^a), for
(n\ne0) |
|
Scientific notation |
(x\times10^y), where
(1\le x<10) |
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