Free Practice Questions for Indices-and-surds in Maths

$\frac{3^{n+2}-3^{n}}{3^{n+1}+3^{n}}=\frac{3^{n} \cdot 9-3^{n}}{3^{n} \cdot 3+3^{n}}=\frac{9-1}{3+1}=\frac{8}{4}=2$

$\begin{aligned}\left(-\frac{1}{512}\right)^{-2 / 3} &=(-512)^{2 / 3} \\ &=\left(-8^{3}\right)^{2 / 3}=(-8)^{2}=64 \end{aligned}$

$3 \sqrt{3} \times 3^{3} \div 3^{-3 / 2}=3^{a+2}$
$\Rightarrow \quad \frac{3 \cdot 3^{1 / 2} \cdot 3^{3}}{3^{-3 / 2}}=3^{a+2}$
$\Rightarrow \quad 3 \cdot 3^{1 / 2} \cdot 3^{3} \cdot 3^{3 / 2}=3^{a+2}$
$\Rightarrow \quad 3^{1+1 / 2+3+3 / 2}=3^{a+2}$
$\Rightarrow \quad 3^{6}=3^{a+2}$
$\Rightarrow \quad 6=a+2 \Rightarrow a=4$

$\begin{aligned} \frac{2^{\frac{2}{3}} \times \sqrt[3]{2^{7}}}{\sqrt[3]{2^{6}}} &=\frac{2^{\frac{2}{3}} \times 2^{\frac{7}{3}}}{2^{\frac{6}{3}}}=2^{\left(\frac{2}{3}+\frac{7}{3}-\frac{6}{3}\right)} \\ &=2^{\left(\frac{2+7-6}{3}\right)}=2^{\frac{3}{3}}=2^{1}=2 \end{aligned}$

$\frac{\left(x^{3}\right)^{2} \times x^{4}}{x^{10}}=x^{p}$
$\Rightarrow \quad \frac{x^{6} \times x^{4}}{x^{10}}=x^{p}$
$\Rightarrow \quad \frac{x^{10}}{x^{10}}=x^{p}$
$\Rightarrow \quad x^{p}=x^{0}$
$\Rightarrow \quad p=0$

$(27)^{1 / 3} \times(81)^{-1 / 2}=3^{n}$
$\Rightarrow \quad\left(3^{3}\right)^{1 / 3} \times\left(3^{4}\right)^{-1 / 2}=3^{n}$
$\Rightarrow \quad 3 \times 3^{-2}=3^{n}$
$\Rightarrow \quad 3^{-1}=3^{n}$
$\Rightarrow \quad n=-1$

$\begin{aligned} \frac{2^{n+4}-2 \cdot 2^{n}}{2 \cdot 2^{n+3}}+2^{-3} &=\frac{2^{n} \cdot 2^{4}-2 \cdot 2^{n}}{2 \cdot 2^{n} \cdot 2^{3}}+\frac{1}{2^{3}} \\ &=\frac{2^{n}(16-2)}{2^{n}(16)}+\frac{1}{8} \\ &=\frac{14}{16}+\frac{1}{8}=\frac{14+2}{16}=\frac{16}{16}=1 \end{aligned}$

$x=7-4 \sqrt{3}$
If the difference of the square of number is 1 then the reciprocal of that number be same with sign change
$\frac{1}{x}=7+4 \sqrt{3}$

$A=\sqrt[3]{2}, \ldots$B$=\sqrt{3}, \ldots C=\sqrt[4]{4}$

$A=(2)^{\frac{1}{3}}, B=(3)^{\frac{1}{2}}, C=(4)^{\frac{1}{4}}$

LCM of $(3,2$ and 4$)=12$

$A=(2)^{\frac{4}{12}}, B=(3)^{\frac{6}{12}}, C=(4)^{\frac{3}{12}}$

$A=\sqrt[12]{2^{4}}, B=\sqrt[12]{3^{6}}, C=\sqrt[12]{4^{3}}$

$A=\sqrt[12]{16}, B=\sqrt[12]{729}, C=\sqrt[12]{64}$

$B>C>A$

$\sqrt[4]{5}, \sqrt[4]{9}, \sqrt[3]{7}, \sqrt{3}$

Take LCM of 4 , and $3=12$ 

$\sqrt[12]{5^{3}}, \sqrt[12]{9^{3}}, \sqrt[12]{7^{4}}, \sqrt[12]{3^{6}}$

The smaller number is $\sqrt[4]{5}$

$3^{2^{x}}=81^{2^{3 x}}$
$3^{2^{x}}=\left(3^{4}\right)^{2^{3 x}}$
$3^{2^{x}}=3^{4 \times 2^{3 x}}$
$3^{2^{x}}=3^{2^{2} \times 2^{3 x}}$
$3^{2^{x}}=3^{2^{2+3x}}$
$ x=2+3 x$
$x=-1$

$3^{a}=9^{b}=27^{c}$
$3^{a}=3^{2 b}=3^{3 c}$
$a=2 b=3 c$
$a=3 c$
$2 b=3 c$
$b=\frac{3 c}{2}$
$a b c=288$
Put the value of $a, b$ in $c$ form
$3 c \times \frac{3 c}{2} \times c=288$
$c^{3}=32 \times 2$
$c=4$
Then $a=3 c=4 \times 3=12$
$
b=\frac{3 c}{2}=\frac{3 \times 4}{2}=6
$
So,
$\frac{1}{3 a}+\frac{1}{5 b}+\frac{1}{7 c}$
$\frac{1}{3 \times 12}+\frac{1}{5 \times 6}+\frac{1}{7 \times 4}$
$\frac{1}{36}+\frac{1}{30}+\frac{1}{28}$
$\frac{35}{1260}+\frac{42}{1260}+\frac{45}{1260}=\frac{122}{1260}=\frac{61}{630}$

$\begin{aligned} &\left(x^{\mathrm{a}-\mathrm{b}}\right)^{l} \times\left(x^{\mathrm{b}-\mathrm{l}}\right)^{\mathrm{a}} \times\left(\mathrm{a}^{l-\mathrm{a}}\right)^{\mathrm{b}} \\=&\left(x^{\mathrm{a}-\mathrm{b}}\right)^{l} \times\left(x^{\mathrm{b}-l}\right)^{\mathrm{a}} \times\left(\mathrm{x}^{l-\mathrm{a}}\right)^{\mathrm{b}} \\=& x^{\mathrm{a} l-\mathrm{b} l} \times x^{\mathrm{ba}-l a} \times x^{\mathrm{lb}-\mathrm{ab}} \\=& x^{\mathrm{a} l-\mathrm{b} l+\mathrm{ba}-\mathrm{la}+\mathrm{lb}-\mathrm{ab}} \\=& x^{0}=1 \end{aligned}$

When a number is written like this and is positive. Then divide the number into two factors such that

In which the difference is only one and also the factor which is larger. That is our answer.

$x=\sqrt{12+\sqrt{12+\sqrt{12+\cdots \infty}}}$
If the number is positive, we take a larger factor.
$=\sqrt{3 \times 4}$ 

4

$\sqrt{5}, \sqrt[3]{3}, \sqrt[3]{9}, \sqrt[3]{4}$

(5) $^{\frac{1}{2}},(3)^{\frac{1}{3}},(9)^{\frac{1}{3}},(4)^{\frac{1}{3}}$

LCM of 2 , and $3=6$

$(5)^{\frac{3}{6}},(3)^{\frac{2}{6}},(9)^{\frac{2}{6}},(4)^{\frac{2}{6}}$

$\sqrt[6]{5^{3}}, \sqrt[6]{3^{2}}, \sqrt[6]{9^{2}}, \sqrt[6]{4^{2}}$

$\sqrt[6]{125}, \sqrt[6]{9}, \sqrt[6]{81}, \sqrt[6]{16}$

The least number is $\sqrt[3]{3}$.

$(36)^{\frac{3}{2}}+(196)^{\frac{1}{2}}$
$(36)^{\frac{3}{2}}+(196)^{\frac{1}{2}}$
$6^{2 \times \frac{3}{2}}+14^{2 \times \frac{1}{2}}$
$216+14$
230

$x=\sqrt[3]{x^{2}+5}-2$
$x+2=\sqrt[3]{x^{2}+5}$
On cubing both sides, we get
$x^{3}+8+6 x^{2}+12 x=x^{2}+5$
$x^{3}+5 x^{2}+12 x=-3$

Let $x=\sqrt{6 \sqrt{6 \sqrt{6 \sqrt{6 \ldots}}}}$
On squaring both sides, we have: $x^{2}=6 x$ Hence $x=6$
Now, $6=(216)^{y-1}$
$1=3 y-3$
$y=4 / 3$

$33-4 \sqrt{35}=33-2 \times 2 \sqrt{5 \times 7}$
$=33-2 \times 2 \sqrt{7} \times \sqrt{5}$
$=28+5-2 \times 2 \sqrt{7} \times \sqrt{5}$
$=(2 \sqrt{7})^{2}+(\sqrt{5})^{2}-2 \times 2 \sqrt{7} \times \sqrt{5}=(2 \sqrt{7}-\sqrt{5})^{2}$
$\therefore \sqrt{33-4 \sqrt{35}}=\sqrt{(2 \sqrt{7}-\sqrt{5})^{2}}=\pm(2 \sqrt{7}-\sqrt{5})$