Question

SOLVE \[\left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7}\]

Answer 1

Hint: Here, all three exponential numbers have the same base .If we take the quotient of two exponentials with the same base, we simply subtract the exponents. If we take the multiplication of two exponentials with the same base, we simply add the exponents.

Formulae used: We use the basic rules of exponentials to solve the above expressions.
1. \[{x^a} \times {x^b} = {x^{a + b}}\]
2.\[{x^a} \div {x^b} = {x^{a - b}}\]
3.\[{\left( {{x^a}} \right)^b} = {x^{ab}}\]
4. BODMAS Rule

Complete step-by-step answer:
\[ \Rightarrow \left[ {{{\left( {{5^2}} \right)}^3} \times {5^4}} \right] \div {5^7}\]
Now we use the formula number 3, and BODMAS we are solving round bracket inside that square bracket where we, consider 5 as ‘x’ and 2 as ‘a’ and 3 as ‘b’, then we get
\[ \Rightarrow \left[ {{5^6} \times {5^4}} \right] \div {5^7}\]
The BODMAS rule states that we should calculate the brackets first then orders, division, multiplication, addition, subtraction respectively. Hence we must solve the bracket and then division is done with\[{5^7}\]. Now inside the bracket we have two exponentials that are \[{5^6}\] & \[{5^4}\] . And there is a sign of multiplication between them. We will use formula of multiplications of exponentials in this case
Using formula number 1 to solve bracket, consider 6 as ‘a’ and 4 as ‘b’ and 5 as ’x’

$ \Rightarrow \left[ {{5^{6 + 4}}} \right] \div {5^7} $
$ \Rightarrow {5^{10}} \div {5^7} $

Now \[{5^{10}}\] and \[{5^7}\] have sign of division in between them, as both are exponentials with same base we will use division formula of exponentials, that is formula number 2
\[ \Rightarrow {5^{10 - 7}}\]
Hence the solution is

\[ \Rightarrow {5^3}\]

Note:
Students must have knowledge of basic rules of exponential and BODMAS rule. Students must check whether expression has the same base or different bases as both the kinds have different rules. Even if they have the same base they have to consider signs between them and brackets.