Question

Question 21
Which of the following is equal to x?

A) $x^{\frac{12}{7}}-x^{\frac{5}{7}}$
B) $\sqrt[12]{12(x^4{)}^{\frac{1}{3}}}$
C) ${\left(\sqrt{x^3}\right)}^{\frac{2}{3}}$
D) $x^{\frac{12}{7}}\times x^{\frac{7}{12}}$

Answer 1

The answer is C.
(a) $x^{\frac{12}{7}}-x^{\frac{5}{7}}=x^{\frac{5}{7}+1}-x^{\frac{5}{7}}$
$=x^{\frac{5}{7}}.x-x^{\frac{5}{7}}[\because a^{m+n}=a^m{a}^n]$
(b) $\sqrt[12]{12(x^4{)}^{\frac{1}{3}}}={\left(\right(x^4{)}^{\frac{1}{3}})}^{\frac{1}{12}}[\sqrt[m]{ma}=a^{a/m}]$
$=x^{4\times \frac{1}{3}\times \frac{1}{12}}=x^{\frac{1}{9}}[\because ((x^m{)}^n{)}^p=x^{mnp}]\ne x$
(c) ${\left(\sqrt{x^3}\right)}^{\frac{2}{3}}=(x^{\frac{3}{2}}{)}^{\frac{2}{3}}=x^{\frac{3}{2}\times \frac{2}{3}}=x[\because (a^m{)}^n=a^{mn}]$
(d) $x^{\frac{12}{7}}\times x^{\frac{7}{12}}=x^{\frac{12}{7}+\frac{7}{12}}[\because a^m.a^n=a^{m+n}]=x^{\frac{144+49}{84}}=x^{\frac{193}{84}}\ne x[\because a^m.a^n=a^{m+n}]$