Question

Prove that :
(i) $\frac{3^{x+1}}{3^{x(x+1)}}\times {\left(\frac{3^x}{3}\right)}^{x+1}=1$
(ii) ${\left(\frac{x^m}{x^m}\right)}^{m+n}\cdot {\left(\frac{x^n}{x^l}\right)}^{n+l}\cdot {\left(\frac{x^l}{x^m}\right)}^{l+m}=1$

Answer 1

(i)$\frac{3^{x+1}}{3^{x(x+1)}}\times {\left(\frac{3^x}{3}\right)}^{x+1}$

$=\frac{3^{x+1}}{3^{x(x+1)}}\times \frac{3^{x(x+1)}}{3^{x+1}}$ ..... [Using power rule of exponents]

$=1$

So LHS=RHS

(ii)${\left(\frac{x^m}{x^n}\right)}^{m+n}.{\left(\frac{x^n}{x^l}\right)}^{n+l}.{\left(\frac{x^l}{x^m}\right)}^{l+m}$

=${\left(x^{m-n}\right)}^{m+n}\times {\left(x^{n-l}\right)}^{n+l}\times {\left(x^{l-m}\right)}^{l+m}$ ...... [Using quotient rule of exponents]

=$\left(x^{m^2-n^2}\right)\times (x{)}^{n^2-l^2}\times (x{)}^{l^2-m^2}$ ...... [Using power rule of exponents]

=$x^{m^2-n^2+n^2-l^2+l^2-m^2}=x^0=1$

So LHS=RHS