Free Exponents and Powers 02 Practice Test - 7th grade 

$81=3\times 3\times 3\times 3=3^4$

$625=5\times 5\times 5\times 5=5^4$

$\therefore \frac{81}{625}=\frac{3^4}{5^4}={\left(\frac{3}{5}\right)}^4$

${(-1)}^{evennumber}=+1$ 

Hence, ${\left(\frac{-3}{5}\right)}^4$  is also correct.

Simplifying  $(\frac{2}{3}{)}^5\times 3^6\times \frac{1}{8}\times \frac{1}{6}$,

we get  $\left(\frac{2^5}{3^5}\right)\times 3^6\times \frac{1}{2^3}\times \frac{1}{2\times 3}$
 = $\frac{2^5\times 3^6}{3^5\times 2^3\times 2\times 3}$

 = $\frac{2^5\times 3^6}{3^6\times 2^4}$

$=2$

Putting $a=2$ and $b=3$,

we get $2^3=2\times 2\times 2=8$.

Here the  two terms  $5^3$ and $2^3$ have different bases, but same exponents.

$5^3\times 2^3=(5\times 5\times 5)\times (2\times 2\times 2)$
$=(5\times 2)\times (5\times 2)\times (5\times 2)$
$=(5\times 2{)}^3$
$={10}^3=1000$
or 
$5^3\times 2^3=(5\times 2{)}^3={10}^3=1000$

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

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

$7^2=49$
$2^2=4$

$7^2\times 2^2=(7\times 2{)}^2={14}^2=196$

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

=$2^9\times 3^6$

$3^3=27$

$2^3=8$

$5^0=1$

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

Here, 10 is called the base.

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

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

Hence, the given statement is false.