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

$0.1234=\frac{1.234}{10}=1.234\times {10}^{-1}$

$512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^9$
$729=3\times 3\times 3\times 3\times 3\times 3=3^6$

=$2^9\times 3^6$

$3^3=27$

$2^3=8$

$5^0=1$

$\Rightarrow 27\times 8\times 1=216$

LHS = $(\frac{2}{5}{)}^2=\frac{4}{25}$

RHS = $-(\frac{2}{5}{)}^2=-\frac{4}{25}$

Hence, the given statement is false.

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

$=3000+20+0.4+0.003$

$=3020.403$

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
$(2^3{)}^2=2^{3\times 2}[\because (x^m{)}^n=x^{(m\times n)}]$
$=2^6$
$=64$
                      

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

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

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

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