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. $(\sqrt[n]{a})^{m} = \sqrt[n]{a^{m}}$

If $\left(\frac{a}{b}\right)^{x-1}=\left(\frac{b}{a}\right)^{x-3}$, then the value of x is:

Answer:

Explanation:

$\left(\frac{a}{b}\right)^{x-1}=\left(\frac{b}{a}\right)^{x-3}$ $ \Leftrightarrow\left(\frac{a}{b}\right)^{x-1}=\left(\frac{a}{b}\right)^{-\left(x-3\right)}$ $=\left(\frac{a}{b}\right)^{3-x}$ $\Leftrightarrow x-1 =3-x$ $\Leftrightarrow 2x=4$ $\Leftrightarrow x=2$

$(25)^{7.5} \times (5)^{2.5}$ $+(125)^{1.5}=5^{?}$

Answer:

Explanation:

Let $(25)^{7.5} \times (5)^{2.5}$ $+(125)^{1.5}=5^{x}$.

Then, $\frac{(25)^{7.5} \times (5)^{2.5}}{(5^{3})^{1.5}} = 5^{x}$

$\Leftrightarrow \frac{5^{(2 \times 7.5 ) \times 5^{2.5}}}{5^(3 \times 1.5)} = 5^{x}$

$\Leftrightarrow \frac{5^{15} \times 5^{2.5}}{5^{4.5}}$

$\Leftrightarrow 5^{x}$
$= 5^{(15+2.5-4.5)} = 5^{13}$
$\Leftrightarrow x = 13$

$(18)^{3.5} +(27)^{3.5}$ $\times 6^{3.5} = 2^{?}$

Answer:

Explanation:

$\Leftrightarrow (18)^{3.5} \times \frac{1}{(27)^{3.5}} \times 6^{3.5}$

$=2^{x}$

$\Leftrightarrow (3^{2} \times 2)^{3.5} \times \frac{1}{(3^{3})^{3.5}} \times (2 \times 3)^{3.5}$

$=2{x}$

$\Leftrightarrow 3^{2 \times 3.5} \times 2{3.5} $ $ \times \frac{1}{3^(3 \times 3.5)} \times 2^{3.5} \times 3^{3.5}$

$=2^{x}$

$\Leftrightarrow 3^{7} \times 2^{3.5}$ $\times \frac{1}{3^{10.5}} \times 2^{3.5} \times 3^{3.5}$

$=2^{x} \Leftrightarrow 2^{7}$

$=2^{x} \Leftrightarrow x = 7$

$(64)^{-\frac{1}{2}}-(-32)^{-\frac{4}{5}}=?$

Answer:

Explanation:

$(64)^{-\frac{1}{2}}-(-32)^{-\frac{4}{5}}$

$=(8^{2})^{-\frac{1}{2}}-((-2)^{5})^{\frac{4}{5}}$

$=8^{2 \times \frac{(-1)}{2}}-(-2)^{5 \times \frac{(-4)}{5}}$

$=8^{-1}-(-2)^{-4}$

$=\frac{1}{8}-\frac{1}{(-2)^{4}}$

$=\left(\frac{1}{8}-\frac{1}{16}\right) = \frac{1}{16}$

$(17)^{3.5} \times (17)^{?}=17^{8}$

Answer:

Explanation:

Let $(17)^{3.5} \times (17)^{x}=17^{8}$.

Then, $(17)^{3.5}+x = (17)^{8}$.

$\therefore$ $3.5+x=8$

$\Leftrightarrow x = (8-3.5)$

$\Leftrightarrow x= 4.5$.

• Surds and Indices - General Questions