Examples (Revised) - Chapter 13 - Exponents & Powers - Ncert Solutions class 7 - Maths

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Chapter 11 - Exponents & Powers - NCERT Solutions Class 7 Maths

Example 1

Express 256 as a power 2.
Solution

We have $256=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$.
So we can say that $256=2^8$

Example 2

Which one is greater $2^3$ or $3^2$ ?
Solution

We have, $2^3=2 \times 2 \times 2=8$ and
$
3^2=3 \times 3=9 \text {. }
$

Since $9>8$, so, $3^2$ is greater than $2^3$
Example 3

Which one is greater $8^2$ or $2^8$ ?
Solution
$
\begin{aligned}
8^2 & =8 \times 8=64 \\
2^8 & =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=256 \\
2^8 & >8^2
\end{aligned}
$
Example 4

Expand $a^3 b^2, a^2 b^3, b^2 a^3, b^3 a^2$. Are they all same?
Solution
$
\begin{aligned}
a^3 b^2 & =a^3 \times b^2 \\
& =(a \times a \times a) \times(b \times b) \\
& =a \times a \times a \times b \times b \\
a^2 b^3 & =a^2 \times b^3 \\
& =a \times a \times b \times b \times b \\
b^2 a^3 & =b^2 \times a^3 \\
& =b \times b \times a \times a \times a \\
b^3 a^2 & =b^3 \times a^2 \\
& =b \times b \times b \times a \times a
\end{aligned}
$

$
=b \times b \times b \times a \times a
$

Note that in the case of terms $a^3 b^2$ and $a^2 b^3$ the powers of $a$ and $b$ are different. Thus $a^3 b^2$ and $a^2 b^3$ are different.

On the other hand, $a^3 b^2$ and $b^2 a^3$ are the same, since the powers of $a$ and $b$ in these two terms are the same. The order of factors does not matter.
Thus, $a^3 b^2=a^3 \times b^2=b^2 \times a^3=b^2 a^3$. Similarly, $a^2 b^3$ and $b^3 a^2$ are the same.
Example 5

Express the following numbers as a product of powers of prime factors:
(i) 72
(ii) 432
(iii) 1000
(iv) 16000

Solution
(i)
$
\begin{aligned}
72 & =2 \times 36=2 \times 2 \times 18 \\
& =2 \times 2 \times 2 \times 9 \\
& =2 \times 2 \times 2 \times 3 \times 3=2^3 \times 3^2
\end{aligned}
$

Thus, $72=2^3 \times 3^2$
(required prime factor product form)

(ii)
$
\begin{aligned}
432=2 \times 216 & =2 \times 2 \times 108=2 \times 2 \times 2 \times 54 \\
& =2 \times 2 \times 2 \times 2 \times 27=2 \times 2 \times 2 \times 2 \times 3 \times 9 \\
& =2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3
\end{aligned}
$
$
\text { or } \quad 432=2^4 \times 3^3
$
(required form)
(iii)
$
\begin{aligned}
1000=2 \times 500 & =2 \times 2 \times 250=2 \times 2 \times 2 \times 125 \\
& =2 \times 2 \times 2 \times 5 \times 25=2 \times 2 \times 2 \times 5 \times 5 \times 5 \\
\text { or } \quad 1000 & =2^3 \times 5^3
\end{aligned}
$

Atul wants to solve this example in another way:
$
\begin{aligned}
1000 & =10 \times 100=10 \times 10 \times 10 \\
& =(2 \times 5) \times(2 \times 5) \times(2 \times 5) \quad(\text { Since } 10=2 \times 5) \\
& =2 \times 5 \times 2 \times 5 \times 2 \times 5=2 \times 2 \times 2 \times 5 \times 5 \times 5 \\
\text { or } \quad 1000 & =2^3 \times 5^3
\end{aligned}
$

Is Atul's method correct?
(iv)
$
\begin{aligned}
& 16,000=16 \times 1000=(2 \times 2 \times 2 \times 2) \times 1000=2^4 \times 10^3(\text { as } 16=2 \times 2 \times 2 \times 2) \\
&=(2 \times 2 \times 2 \times 2) \times(2 \times 2 \times 2 \times 5 \times 5 \times 5)=2^4 \times 2^3 \times 5^3 \\
&\quad \text { Since } 1000=2 \times 2 \times 2 \times 5 \times 5 \times 5) \\
&=(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2) \times(5 \times 5 \times 5) \\
& \text { or, } 16,000= 2^7 \times 5^3
\end{aligned}
$

Example 6

Work out $(1)^5,(-1)^3,(-1)^4,(-10)^3,(-5)^4$.
Solution
(i) We have $(1)^5=1 \times 1 \times 1 \times 1 \times 1=1$
In fact, you will realise that 1 raised to any power is 1 .
(ii) $(-1)^3=(-1) \times(-1) \times(-1)=1 \times(-1)=-1$
(iii) $(-1)^4=(-1) \times(-1) \times(-1) \times(-1)=1 \times 1=1$

You may check that ( -1 ) raised to any odd power is (-1), and $(-1)$ raised to any even power is $(+1)$.
(iv) $(-10)^3=(-10) \times(-10) \times(-10)=100 \times(-10)=-1000$
(v) $(-5)^4=(-5) \times(-5) \times(-5) \times(-5)=25 \times 25=625$

Example 7

Can you tell which one is greater $\left(5^2\right) \times 3$ or $\left(5^2\right)^3$ ?
Solution

$\quad\left(5^2\right) \times 3$ means $5^2$ is multiplied by 3 i.e., $5 \times 5 \times 3=75$
but $\left(5^2\right)^3$ means $5^2$ is multiplied by itself three times i.e.,
$
5^2 \times 5^2 \times 5^2=5^6=15,625
$

Therefore
$
\left(5^2\right)^3>\left(5^2\right) \times 3
$

Example 8

Express the following terms in the exponential form:
(i) $(2 \times 3)^5$
(ii) $(2 a)^4$
(iii) $(-4 m)^3$

Solution
(i)
$
\begin{aligned}
(2 \times 3)^5 & =(2 \times 3) \times(2 \times 3) \times(2 \times 3) \times(2 \times 3) \times(2 \times 3) \\
& =(2 \times 2 \times 2 \times 2 \times 2) \times(3 \times 3 \times 3 \times 3 \times 3) \\
& =2^5 \times 3^5
\end{aligned}
$

(ii)
$
\begin{aligned}
(2 a)^4 & =2 a \times 2 a \times 2 a \times 2 a \\
& =(2 \times 2 \times 2 \times 2) \times(a \times a \times a \times a) \\
& =2^4 \times a^4
\end{aligned}
$
(iii)
$
\begin{aligned}
(-4 m)^3 & =(-4 \times m)^3 \\
& =(-4 \times m) \times(-4 \times m) \times(-4 \times m) \\
& =(-4) \times(-4) \times(-4) \times(m \times m \times m)=(-4)^3 \times(m)^3
\end{aligned}
$

Example 9

Expand:
(i) $\left(\frac{3}{5}\right)^4$
(ii) $\left(\frac{-4}{7}\right)^5$

Solution
(i) $\left(\frac{3}{5}\right)^4=\frac{3^4}{5^4}=\frac{3 \times 3 \times 3 \times 3}{5 \times 5 \times 5 \times 5}$
(ii) $\left(\frac{-4}{7}\right)^5=\frac{(-4)^5}{7^5}=\frac{(-4) \times(-4) \times(-4) \times(-4) \times(-4)}{7 \times 7 \times 7 \times 7 \times 7}$

Example 10

Write exponential form for $8 \times 8 \times 8 \times 8$ taking base as 2 .
Solution

We have, $8 \times 8 \times 8 \times 8=8^4$
But we know that
$
\begin{aligned}
8 & =2 \times 2 \times 2=2^3 \\
8^4 & =\left(2^3\right)^4=2^3 \times 2^3 \times 2^3 \times 2^3 \\
& =2^{3 \times 4} \quad\left[\text { You may also use }\left(a^m\right)^n=a^{n m}\right] \\
& =2^{12} \quad
\end{aligned}
$

Example 11

Simplify and write the answer in the exponential form.
(i) $\left(\frac{3^7}{3^2}\right) \times 3^5$
(ii) $2^3 \times 2^2 \times 5^5$
(iii) $\left(6^2 \times 6^4\right) \div 6^3$
(iv) $\left[\left(2^2\right)^3 \times 3^6\right] \times 5^6$
(v) $8^2 \div 2^3$

Solution
$
\text { (i) } \begin{aligned}
\left(\frac{3^7}{3^2}\right) \times 3^5 & =\left(3^{7-2}\right) \times 3^5 \\
& =3^5 \times 3^5=3^{5+5}=3^{10}
\end{aligned}
$

(ii)
$
\begin{aligned}
2^3 \times 2^2 \times 5^5 & =2^{3+2} \times 5^5 \\
& =2^5 \times 5^5=(2 \times 5)^5=10^5
\end{aligned}
$
(iii)
$
\begin{aligned}
\left(6^2 \times 6^4\right) \div 6^3 & =6^{2+4} \div 6^3 \\
& =\frac{6^6}{6^3}=6^{6-3}=6^3
\end{aligned}
$
(iv)
$
\begin{aligned}
{\left[\left(2^2\right)^3 \times 3^6\right] \times 5^6 } & =\left[2^6 \times 3^6\right] \times 5^6 \\
& =(2 \times 3)^6 \times 5^6 \\
& =(2 \times 3 \times 5)^6=30^6
\end{aligned}
$
(v) $8=2 \times 2 \times 2=2^3$

Therefore $8^2 \div 2^3=\left(2^3\right)^2 \div 2^3$
$
=2^6 \div 2^3=2^{6-3}=2^3
$

Example 12

Simplify:
(i) $\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27}$
(ii) $2^3 \times a^3 \times 5 a^4$
(iii) $\frac{2 \times 3^4 \times 2^5}{9 \times 4^2}$

Solution
(i) We have

$\begin{aligned}
\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27} & =\frac{\left(2^2 \times 3\right)^4 \times\left(3^2\right)^3 \times 2^2}{(2 \times 3)^3 \times\left(2^3\right)^2 \times 3^3} \\
& =\frac{\left(2^2\right)^4 \times(3)^4 \times 3^{2 \times 3} \times 2^2}{2^3 \times 3^3 \times 2^{2 \times 3} \times 3^3}=\frac{2^8 \times 2^2 \times 3^4 \times 3^6}{2^3 \times 2^6 \times 3^3 \times 3^3} \\
& =\frac{2^{8+2} \times 3^{4+6}}{2^{3+6} \times 3^{3+3}}=\frac{2^{10} \times 3^{10}}{2^9 \times 3^6} \\
& =2^{10-9} \times 3^{10-6}=2^1 \times 3^4 \\
& =2 \times 81=162
\end{aligned}$

$\text { (ii) } \begin{aligned}
2^3 \times a^3 \times 5 a^4 & =2^3 \times a^3 \times 5 \times a^4 \\
& =2^3 \times 5 \times a^3 \times a^4=8 \times 5 \times a^{3+4} \\
& =40 a^7
\end{aligned}$

(ii) $\frac{2 \times 3^4 \times 2^5}{9 \times 4^2}=\frac{2 \times 3^4 \times 2^5}{3^2 \times\left(2^2\right)^2}=\frac{2 \times 2^5 \times 3^4}{3^2 \times 2^{2 \times 2}}$
$
\begin{aligned}
& =\frac{2^{1+5} \times 3^4}{2^4 \times 3^2}=\frac{2^6 \times 3^4}{2^4 \times 3^2}=2^{6-4} \times 3^{4-2} \\
& =2^2 \times 3^2=4 \times 9=36
\end{aligned}
$

Note: In most of the examples that we have taken in this Chapter, the base of a power was taken an integer. But all the results of the chapter apply equally well to a base which is a rational number.

Example 13

Express the following numbers in the standard form:
(i) 5985.3
(ii) 65,950
(iii) $3,430,000$
(iv) $70,040,000,000$

Solution
(i) $5985.3=5.9853 \times 1000=5.9853 \times 10^3$
(ii) $65,950=6.595 \times 10,000=6.595 \times 10^4$
(iii) $3,430,000=3.43 \times 1,000,000=3.43 \times 10^6$
(iv) $70,040,000,000=7.004 \times 10,000,000,000=7.004 \times 10^{10}$

A point to remember is that one less than the digit count (number of digits) to the left of the decimal point in a given number is the exponent of 10 in the standard form. Thus, in $70,040,000,000$ there is no decimal point shown; we assume it to be at the (right) end. From there, the count of the places (digits) to the left is 11 . The exponent of 10 in the standard form is $11-1=10$. In 5985.3 there are 4 digits to the left of the decimal point and hence the exponent of 10 in the standard form is $4-1=3$.