Free Exponents and Powers 01 Practice Test - 7th grade 

$3\times 1000+2\times 10+\frac{4}{10}+\frac{3}{1000}$

$=3000+20+0.4+0.003$

$=3020.403$

We can write:
$81=3\times 3\times 3\times 3$
$16=2\times 2\times 2\times 2$

So, $\frac{81}{16}$ = $\left(\frac{3^4}{2^4}\right)$

And we know that, $\left(\frac{a^m}{b^m}\right)=(\frac{a}{b}{)}^m$
$\therefore \frac{81}{16}$ = $\left(\frac{3^4}{2^4}\right)$ $=(\frac{3}{2}{)}^4$

Third power of $\frac{1}{4}=(\frac{1}{4}{)}^3$
Fourth power of  4 is equal to $4^4$.

$(\frac{1}{4}{)}^3\times 4^4$
= $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times 4\times 4\times 4\times 4$
= $\frac{4\times 4\times 4\times 4}{4\times 4\times 4}$
= $\frac{4^4}{4^3}$
= 4

On substituting a = 2 and b = 3, we get

$(2^3{)}^2=2^{3\times 2}[\because (x^m{)}^n=x^{(m\times n)}]$

$=2^6$

$=64$
                      

Notice that here the two terms $4^3$ and $2^3$ have different bases, but the same exponents.

$4^3\times 2^3=(4\times 4\times 4)\times (2\times 2\times 2)$
$=(4\times 2)\times (4\times 2)\times (4\times 2)=8^3=512$

In standard form, any number is expressed in powers of 10. The decimal number must be in between 1 and 10.

$5.126\times 1000000=5.126\times {10}^6$

We can write:
$256=(2\times 2\times 2\times 2\times 2\times 2\times 2\times 2)=2^8$ 

$243=(3\times 3\times 3\times 3\times 3)=3^5$

$\therefore 256\times 243=2^8\times 3^5$

$\frac{a^b}{a^c}=a^{b-c}$      [a is a non-zero integer]

Using this law, we get,

$(\frac{-2}{5}{)}^7÷(\frac{-2}{5}{)}^6$

${\left(\frac{-2}{5}\right)}^{7-6}={\left(\frac{-2}{5}\right)}^1=\frac{-2}{5}$

$2^3\times 3^3=(2\times 3{)}^3=6^3=216$

$\because$ For any number a $\ne$ 0, a${}^0$ = 1.

$\therefore 3^0=2^0=1$ 

$\Rightarrow (3^0-2^0)\times 5^2=(1-1)\times 25=0$