Question

The Multiplicative inverse of 2m is _______
36/121
121/36
72/121
121/72

Answer 1

given,

$m=\left(\left(\frac{2}{3}\right)-\left(\frac{2}{5}\right)^{-1}\right)^{-2}\\m=\left(\frac{2}{3}-\left(\frac{2}{5}\right)^{-1}\right)^{-2}\\\left(\frac{2}{5}\right)^{-1} =\frac{1}{\frac{2}{5}}\\=\frac{5}{2}\\=\frac{2}{3}-\frac{5}{2}\\=\frac{4}{6}-\frac{15}{6}\\=\frac{4-15}{6}\\=\frac{-11}{6}\\=-\frac{11}{6}\\\mathrm{Multiply\:each\:numerator\:by\:the\:same\:amount\:needed\:to\:multiply\:its}\\\mathrm{corresponding\:denominator\:to\:turn\:it\:into\:the\:LCM}\:6$

$\frac{2}{3}=\frac{2\cdot \:2}{3\cdot \:2}\\=\frac{4}{6}\\\\\frac{5}{2}=\frac{5\cdot \:3}{2\cdot \:3}\\=\frac{15}{6}\\=\frac{4}{6}-\frac{15}{6}\\=\frac{-11}{6}\\=-\frac{11}{6}\\m=\left(-\frac{11}{6}\right)^{-2}\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}\\m=\frac{1}{\left(-\frac{11}{6}\right)^2}\\\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^n=a^n,\:\mathrm{if\:}n\mathrm{\:is\:even}\\\left(-\frac{11}{6}\right)^2=\left(\frac{11}{6}\right)^2$

$m=\frac{36}{11^2}\\11^2=121\\m=\frac{36}{121}$

Hence, the correct option is 36/121   .