R S Aggarwal and Veena Aggarwal Solutions for Exercise 7: EXERCISE 1G

Find the value of $x$ in the following.

$\frac{3^{3x}·3^{2x}}{3^x}=\sqrt[4]{3^{20}}$

Prove that

$\sqrt{x^{-1}y}·\sqrt{y^{-1}z}·\sqrt{z^{-1}x}=1$

Prove that ${\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$

Prove that

$\frac{x^{a(b-c)}}{x^{b(a-c)}}÷{\left(\frac{x^b}{x^a}\right)}^c=1$

Prove that $\frac{{\left(x^{a+b}\right)}^2{\left(x^{b+c}\right)}^2{\left(x^{c+a}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}=1$

If $x$ is a positive real number and exponents are rational numbers, simplify

${\left(\frac{x^b}{x^c}\right)}^{b+c-a}·{\left(\frac{x^c}{x^a}\right)}^{c+a-b}·{\left(\frac{x^a}{x^b}\right)}^{a+b-c}$

If $\frac{9^n\times 3^2\times {\left(3^{-\frac{n}{2}}\right)}^{-2}-(27{)}^n}{3^{3m}\times 2^3}=\frac{1}{27}$, Prove that $m-n=1$.

Choose the correct option for the following in ascending order of magnitude.
$\sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8}$

$$\begin{gathered} A: \sqrt[6]{6}<\sqrt[4]{8}<\sqrt[3]{7} \\ B: \sqrt[4]{8}<\sqrt[3]{7}<\sqrt[6]{6} \\ C: \sqrt[6]{6}<\sqrt[3]{7}<\sqrt[4]{8} \end{gathered}$$