Arithmetic aptitude :: Surds and Indices :: General Questions

• Surds and Indices - General Questions
• Surds and Indices - Important Formulas and Facts

Laws of Indices :

1.  $a^{m} \times a^{n}= a^{m+n}$
2.  $\frac{a^{m}}{a^{n}}= a^{m-n}$
3.  $(a^{m})^{n}=a^{mn}$
4.  $(ab)^{n} = a^{n}b^{n}$
5.  $\left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}}$
6.  $a^{0}=1$

Surds:

When we can't simplify a number to remove a square root (or cube root etc) then it is called a surd.

Rational Number:

A real number that can be written as a simple fraction then it is called rational number

e.g. 1.5 = $\frac{3}{2}$

Irrational Number:

A real number that can't be written as a simple fraction is called Irrational number.

e.g. $\pi$ = 3.1415926535897932384626433832795

Let a be rational number and n be a positive integer such that $a^{\frac{1}{n}}=\sqrt[n]{a}$ is irrational. Then $\sqrt[n]{a}$ is called surd of order n.

Laws of Surds

1. $\sqrt[n]{a}=a^{\frac{1}{n}}$
2. $\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$
3. $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$
4. $(\sqrt[n]{a})^{n}= a$
5. $\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$
6. $(\sqrt[n]{a})^{m} = \sqrt[n]{a^{m}}$

Which is larget $\sqrt{2}$ or $\sqrt[3]{3}$?

Answer:

Explanation:

Given surds are of order 2 and 3, Their LCM is 6.

Changing each to a surd of order 6, we get :

$\sqrt{2} = 2^{\frac{1}{2}}$
$=2^{\left(\frac{1}{2} \times \frac{3}{3}\right)}$
$=2^{\frac{3}{6}}$
$=(2^{3})^{\frac{1}{6}}$
$=(8)^{\frac{1}{6}}$
$=\sqrt[6]{8}$

$\sqrt[3]{3} = 3^{\frac{1}{3}}$
$=3^{\left(\frac{1}{3} \times \frac{2}{2}\right)}$
$=3^{\frac{2}{6}}$
$=(3^{2})^{\frac{1}{6}}$
$=(9)^{\frac{1}{6}}$
$=\sqrt[6]{9}$
Clearly, $\sqrt[6]{9} > \sqrt[6]{8}$ and hence $\sqrt[3]{3}> \sqrt{2}$

If $2^{x}=3^{y}=6^{-z}$, then $\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is equal to?

Answer:

Explanation:

Let $2^{x}=3^{y}=6^{-z}=k$ $\Leftrightarrow 2=k^{\frac{1}{x}},$ $\Leftrightarrow 3=k^{\frac{1}{y}} $ and $\Leftrightarrow 6=k^{\frac{-1}{z}}.$ Now $2\times 3=6$ $\Leftrightarrow k^{\frac{1}{x}} \times k^{\frac{1}{y}}=k^{\frac{- 1}{z}}$ $\Leftrightarrow k^{\left(\frac{1}{x} +\frac{1}{y} \right)}=k^{\frac{- 1}{z}}$ $\therefore \frac{1}{x}+\frac{1}{y}=\frac{-1}{z}$ $\Leftrightarrow \frac{1}{x} + \frac{1}{y}+\frac{1}{z}=0$

If $2^{x}=\sqrt[3]{32}$ , then x is equal to?.

Answer:

Explanation:

$2^{x}=\sqrt[3]{32} \Leftrightarrow 2^{x}=32^{\frac{1}{3}}$ $=\left(2^{5}\right)^{\frac{1}{3}} =2^{\frac{5}{3}}$ $\Leftrightarrow x=\frac{5}{3}$

If  $3^{(x-y)}=27 $  and $3^{(x+y)}=243 $, then x is equal to?

Answer:

Explanation:

$3^{x-y}=27=3^{3}$ $\Leftrightarrow x-y=3$ $3^{x+y}=243=3^{5}$ $\Leftrightarrow x+y=5$ On solving these equations, we get x = 4.

if abc= 1, then $\left(\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}\right)=?$

Answer:

Explanation:

Given Exp. $=\left(\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}\right)$ $=\frac{1}{1+a+b^{-1}}+\frac{b^{-1}}{b^{-1}+1+b^{-1}c^{-1}}+\frac{a}{a+ac+1}$ $=\frac{1}{1+a+b^{-1}}+\frac{b^{-1}}{1+b^{-1}+a}+\frac{a}{a+b^{-1}+1}$ $=\frac{1+a+b^{-1}}{1+a+b^{-1}}=1$ $\left[\because abc=1 \right] \Rightarrow \left(bc\right)^{-1}$ $=a\Rightarrow b^{-1}c^{-1}=a$ and $ac= b^{-1}$

• Surds and Indices - General Questions