Question

If $(2^{-1}\times 4^{-1})÷2^{-2}$ is $\frac{1}{m}$, then find the value of $m$.

Answer 1

$(2^{-1}\times 4^{-1})÷2^{-2}$

$=\frac{(2^{-1}\times 4^{-1})}{2^{-2}}$

By laws of indices

$m^{-1}=\frac{1}{m^1}$

$=\frac{(\frac{1}{2}\times \frac{1}{4})}{{\left(\frac{1}{2}\right)}^2}$ $=\frac{\left(\frac{1}{8}\right)}{\left(\frac{1}{4}\right)}$ $=\frac{4}{8}$ $=\frac{1}{2}$

Thus, $m=2$