Question

express (4^-3) in power notation,with base 2
CLASS-8​
PLEASE HELP

Answer 1

Required Answer:

Solution:

$\sf { \longrightarrow ({4}^{-3})}$

$\sf { \longrightarrow ({2 \times 2}^{-3})}$

$\sf { \longrightarrow ({{2}^{2}}^{-3})}$

$\sf { \longrightarrow [{2}^{2 \times (-3)}]}$

[Since,$\sf{ {{a}^{b}}^{c}= {a}^{b \times c} }$

$\sf\purple { \longrightarrow {2}^{-6}}$

__________________________

How did I solve this?

Here, as per the given question, we have to express $\sf { ({4}^{-3})}$ in power notation, with the base 2.

So, firstly we will break the given base (4) into number. We can write 4 as 2 × 2.

$\sf { \longrightarrow {4}^{-3} = {(2 \times 2)}^{-3}}$

Also we can write 2 × 2 as 2².

$\sf { \longrightarrow {4}^{-3} = {({2}^{2})}^{-3}}$

Now, by using one of the laws of exponents , that is,

• $\sf { {({a}^{b})}^{c} = {a}^{b \times c}}$

$\sf { \longrightarrow [{2}^{2 \times (-3)}]}$

Now , performing multiplication:

$\sf\purple { \longrightarrow {2}^{-6}}$.

So, the answer is $\sf { {2}^{-6}}$.

_______________________________

Extra:

Laws of exponents:

• $\sf { {a}^{0} = 1}$

• $\sf { {a}^{m} \times {a}^{n} = {a}^{m+n} }$

• $\sf { {a}^{m} \times {b}^{m} = {(ab)}^{m} }$

• $\sf { {a}^{-m} = \dfrac{1}{{a}^{m}} }$

• $\sf { \dfrac{{a}^{m}}{{a}^{n}} = {a}^{m-n} }$

_______________________________