Free Practice Questions for Indices-and-surds in Maths

$(6 \times 6 \times 6 \times 6 \times 6 \times 6)^{7} \times(6 \times 6 \times 6)^{5} \div(6)^{3}$
$=(216)$ ?
$\Rightarrow \quad\left(6^{6}\right)^{7} \times\left(6^{3}\right)^{5} \div(6)^{3}=(6)^{3} ?$
$\Rightarrow \quad 6^{42+15-3}=\left(6^{3}\right) ?$
$\Rightarrow \quad 6^{54}=\left(6^{3}\right) ?$
$\Rightarrow \quad\left(6^{3}\right)^{18}=\left(6^{3}\right) ?$
$?=18$

Since, 

The surds of $\sqrt[3]{4}, \sqrt{2}, \sqrt[6]{3}, \sqrt[4]{5}$  are $3,2,6$ and 4 respectively. 

Therefore, the L.C.M is 12.
$\begin{aligned}
&\sqrt[3]{4}=\sqrt[3 \times 4]{4^{4}}=\sqrt[12]{256} \\
&\sqrt{2}=\sqrt[2 \times 6]{4^{6}}=\sqrt[12]{64} \\
&\sqrt[6]{3}=\sqrt[6 \times 2]{3^{2}}=\sqrt[12]{9} \\
&\sqrt[4]{5}=\sqrt[4 \times 3]{5^{3}}=\sqrt[12]{125}\end{aligned}$
In decreasing order,
$\begin{aligned}\sqrt[12]{256} &>\sqrt[12]{125}>\sqrt[12]{64}>\sqrt[12]{9} \\
=& \sqrt[3]{4}>\sqrt[4]{5}>\sqrt{2}>\sqrt[6]{3}
\end{aligned}$

$\left(\frac{1}{216}\right)^{-\frac{2}{3}} \div\left(\frac{1}{27}\right)^{-\frac{4}{3}}=x$
$\Rightarrow \quad\left(\frac{1}{6^{3}}\right)^{-\frac{2}{3}} \div\left(\frac{1}{3^{3}}\right)^{-\frac{4}{3}}=x$
$\Rightarrow \quad\left(\frac{1}{6^{-2}}\right) \div\left(\frac{1}{3^{-4}}\right)=x$
$\Rightarrow \quad 6^{2} \div 3^{4}=x$
$\Rightarrow \quad \frac{6 \times 6}{3 \times 3 \times 3 \times 3}=x$
$x=\frac{4}{9}$

$\left[3^{m^{2}} \div\left(3^{m}\right)^{2}\right]^{1 / m}=81 \Rightarrow\left[3^{m^{2}} \div 3^{2 m}\right]^{1 / m}=3^{4}$
$
\begin{aligned}
&\left(\frac{3^{m^{2}}}{3^{2 m}}\right)^{\frac{1}{m}}=3^{4} \\
\Rightarrow \quad\left(3^{m^{2}-2 m}\right)^{\frac{1}{m}} &=3^{4}
\end{aligned}
$
\begin{aligned}
&\Rightarrow \quad\left(3^{m^{2}-2 m}\right)^{\frac{1}{m}}=3^{4} \\
&\Rightarrow \quad 3 \frac{m(m-2)}{m}=3^{4} \Rightarrow 3^{m-2}=3^{4} \\
&\Rightarrow \quad m-2=4 \quad \Rightarrow \quad m=4+2=6
\end{aligned}

$\left(\frac{x^{a}}{x^{b}}\right)^{1 / a b} \times\left(\frac{x^{b}}{x^{c}}\right)^{1 / b c} \times\left(\frac{x^{c}}{x^{a}}\right)^{1 / a c}$
$=\left(\frac{x^{a / a b}}{x^{b / a b}}\right) \times\left(\frac{x^{b / b c}}{x^{c / b c}}\right) \times\left(\frac{x^{c / a c}}{x^{a / a c}}\right)$
$=\left(\frac{x^{\frac{1}{b}}}{x^{\frac{1}{a}}}\right) \times\left(\frac{x^{\frac{1}{c}}}{x^{\frac{1}{b}}}\right) \times\left(\frac{x^{\frac{1}{a}}}{x^{\frac{1}{c}}}\right)$
$=x^{\frac{1}{b}-\frac{1}{a}} \times x^{\frac{1}{c}-\frac{1}{b}} \times x^{\frac{1}{a}-\frac{1}{c}}=x^{\frac{1}{b}-\frac{1}{a}+\frac{1}{c}-\frac{1}{b}+\frac{1}{a}-\frac{1}{c}=x^{0}=1} $

$4^{x}-4^{x-1}=24$
$\Rightarrow \quad 4^{x}-4^{x} \cdot 4^{-1}=24$
$\Rightarrow \quad 4^{x}\left(1-\frac{1}{4}\right)=24 \quad \Rightarrow \quad 4^{x}\left(\frac{3}{4}\right)=24$
$\Rightarrow \quad 4^{x}=\frac{24 \times 4}{3} \Rightarrow 4^{x}=2^{5}$
$\Rightarrow \quad 2^{2 x}=2^{5} \Rightarrow 2 x=5$
$\Rightarrow \quad x=\frac{5}{2}$
अब, $\quad(2 x)^{x}=(5)^{5 / 2}=25 \sqrt{5}$

$(900)^{1 / 2} \times(0.027)^{1 / 3}-(0.0081)^{1 / 4} \times 3^{0}+\left(\frac{5}{4}\right)^{-1}$
$=30 \times 0.3-0.3+\frac{4}{5}=9-0.3+0.8=9.5$

$\sqrt{2^{4}}+\sqrt[3]{64}+\sqrt[4]{2^{8}}$
$\quad=2^{2}+4+2^{2}=4+4+4=12$

$\frac{\left(3^{2}\right)^{n} \times 3^{5} \times\left(3^{3}\right)^{3}}{3 \times\left(3^{4}\right)^{4}}=3^{3}$
$\Rightarrow \quad \frac{3^{2 n} \times 3^{5} \times 3^{9}}{3 \times 3^{16}}=3^{3}$
$\Rightarrow 3^{2 n+5+9-1-16}=3^{3}$
$\Rightarrow \quad 2 n-3=3$
$\Rightarrow \quad n=3$

माना $a^{x}=b^{y}=c^{z}=k$
$\Rightarrow a=k^{1 / x}, b=k^{1 / y}$ तथा $c=k^{1 / z}$
तथा $b^{2}=a c$
$\Rightarrow \quad k^{2 / y}=k^{1 / x} \cdot k^{1 / z}$
$\Rightarrow \quad k^{2 / y}=k^{1 / x+1 / z} \Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z}$
$\Rightarrow \quad \frac{2}{y}=\frac{z+x}{x z} \quad \Rightarrow \quad y=\frac{2 x z}{x+z}$

$\sqrt{5+\sqrt[3]{x}}=3$
$\Rightarrow 5+\sqrt[3]{x}=9$
$\Rightarrow \quad \sqrt[3]{x}=4$
$\Rightarrow \quad x=64$