Question

Evaluate each of the following
I. \[{5^2} \times {5^4}\]
II. \[{5^8} + {5^3}\]
III. \[{\left( {{3^2}} \right)^4}\]
IV. \[{\left( {\dfrac{{11}}{{12}}} \right)^2}\]

Answer 1

Hint: First we have to know what surds and indices. Mention their laws. Then consider each option separately using laws of surds and indices evaluate it.

Complete step-by-step answer:
Laws of indices: Suppose \[a\] and \[b\] are any two non-zero real numbers. Let \[n\] and \[m\] be any two non-zero integers then the following laws holds:
 \[{a^m} \times {a^n} = {a^{m + n}}\]

 \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
 \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
 \[{\left( {ab} \right)^n} = {a^n}{b^n}\]
 \[{\left( {\dfrac{a}{b}} \right)^n} = \dfrac{{{a^n}}}{{{b^n}}}\]
 \[{a^0} = 1\]

Option (i):
Given \[{5^2} \times {5^4}\] ----(1)
Using the first law of indices in the expression (1), we get
 \[{5^2} \times {5^4} = {5^{2 + 4}} = {5^6} = 15625\] .
Hence the value of \[{5^2} \times {5^4}\] is \[15625\] .

Option (ii):
Given \[{5^8} + {5^3}\] ----(4)
Take \[{5^3}\] common in each term in the expression (2), we get
 \[{5^3}({5^5} + 1) = 125 \times 3126 = 390750\] .
Hence the value of \[{5^8} + {5^3}\] is \[390750\] .

Option (iii):
Given \[{\left( {{3^2}} \right)^4}\] ----(3)
Using the third law of indices in the expression (3), we get
 \[{\left( {{3^2}} \right)^4} = {3^8} = 81 \times 81 = 6561\] .
Hence the value of \[{\left( {{3^2}} \right)^4}\] is \[6561\] .

Option (vi):
Given \[{\left( {\dfrac{{11}}{{12}}} \right)^2}\] ----(4)
Using the fifth law of indices in the expression (3), we get
 \[{\left( {\dfrac{{11}}{{12}}} \right)^2} = \dfrac{{{{11}^2}}}{{{{12}^2}}} = 0.8402\bar 7\] .

Note: Let \[a\] be a rational number and \[n\] be a positive integer such that \[{a^{\dfrac{1}{n}}} = \sqrt[n] {a}\] is irrational. Then, \[\sqrt[n] {a}\] is called \[a\] surd of order \[n\] .
Laws of surds: Suppose \[a\] and \[b\] are any two non-zero real numbers. Let \[n\] and \[m\] be any two non-zero integers then the following laws holds:
I. \[\sqrt[n] {a} = {a^{\dfrac{1}{n}}}\]
II. \[\sqrt[n] {{ab}} = {a^{\dfrac{1}{n}}} \times {b^{\dfrac{1}{n}}}\]
III. \[\sqrt[n] {{\dfrac{a}{b}}} = \dfrac{{\sqrt[n] {a}}}{{\sqrt[n] {b}}}\]
IV. \[\left( {\sqrt[n] {a}} \right) = a\]
V. \[\sqrt[m] {{\left( {\sqrt[n] {a}} \right)}} = \sqrt[{mn}] {a}\]
VI. \[{\left( {\sqrt[m] {a}} \right)^n} = \sqrt[m] {{\left( {{a^n}} \right)}}\]