Free Exponents and Powers 03 Practice Test - 8th Grade 

${(2^5÷2^8)}^5\times 2^{-5}={\left(\frac{2^5}{2^8}\right)}^5\times 2^{-5}$
$={\left(2^{5-8}\right)}^5\times 2^{-5}(\because \frac{a^m}{a^n}=a^{m-n})$
$={\left(2^{-3}\right)}^5\times 2^{-5}$
$=2^{-15}\times 2^{-5}(\because (a^m{)}^n=a^{mn})$
$=2^{-15-5}(\because a^m\times a^n=a^{m+n})$
$=2^{-20}$
$=\frac{1}{2^{20}}(\because a^{-m}=\frac{1}{a^m})$

Note that $a^0=1$ for any non-zero integer $a$.
Also, for integers $a(\ne 0)$ and $m,$
$a^{-m}=\frac{1}{a^m}.$
$⟹4^{-2}=\frac{1}{4^2}=\frac{1}{16}$ and $4^{-1}=\frac{1}{4^1}$
 $\therefore 4^{-2}+4^{-1}+4^0+4^2=\frac{1}{16}+\frac{1}{4}+1+16$
$=\frac{1+4+16+256}{16}$
$=\frac{277}{16}$

The value of ($4^{-7}$ ÷ $4^{-10}$ ) × $4^{-5}$ is ${16}^{-1}$.
$(4^{-7}$÷ $4^{-10})\times 4^{-5}$

=$\left(\frac{4^{-7}}{4^{-10}}\right)\times 4^{-5}$
=$4^{-7+10}\times 4^{-5}(\because \frac{a^m}{a^n}=a^{m-n})$
$=4^3\times 4^{-5}$
$\Rightarrow 4^{3-5}(\because a^m\times a^n=a^{m+n})$
$=4^{-2}=\frac{1}{4^2}$
$=\frac{1}{16}$
$={16}^{-1}$

$(3ab{)}^2$$(-5a^{2}b{c}^4{)}^2$
= (9$a^2$$b^2$) × (25$a^4$$b^2$$c^8$)
= 225$a^6$$b^4$$c^8$
Exponent of c = 8

We know that for a non-zero integer $a$ and any integer $m,$
$a^{-m}=\frac{1}{a^m}.$
$\therefore 6^{-2}=\frac{1}{6^2}$
$⟹\frac{1}{6^{-2}}=6^2=36$
Similarly, $\frac{1}{3^{-3}}=3^3=27$ and $\frac{1}{5^{-2}}=5^2=25.$
$⟹(\frac{1}{6^{-2}}-\frac{1}{3^{-3}})÷\frac{1}{5^{-2}}=(36-27)÷25$
$=\frac{9}{25}$
$=\frac{3^2}{5^2}$
$=\left(\frac{3}{5}{)}^2\right(\because \frac{a^m}{b^m}=(\frac{a}{b}{)}^m$

$8,780,000,000,000=878\times {10}^{10}$
$=8.78\times {10}^2\times {10}^{10}$
$=8.78\times {10}^{12}$
$=87.8\times {10}^{-1}\times {10}^{12}$
$=87.8\times {10}^{11}$
(Using $a^m\times a^n=a^{m+n}$)

1 micron = ${10}^{-6}$ m.

$2^7$ is $2$ multiplied by itself, 7 times.

So, $2^7=2\times 2\times 2\times 2\times 2\times 2\times 2=128$

${\left(\right(2{)}^2)}^2$× ($3^3$ ÷$4^2$) ÷ $3^2$
= $4^2$ ×$\frac{27}{16}$× $\frac{1}{9}(\because (a^m{)}^n=a^{mn})$
= 16 × $\frac{27}{16}$× $\frac{1}{9}$ 
= $\frac{27}{9}$
= 3

$2^{10}$= $(2^5{)}^2$ = $(2\times 2\times 2\times 2\times 2{)}^2$ = 32×32 = 1024

${10}^2$ = 10×10 = 100

Since 1024 > 100 so, $2^{10}$ is greater than  ${10}^2$.