Question

find the value of
$({3}^{ - 1} + {4}^{ - 2} + {6}^{ - 4} ) ^{ - 1}$

Answer 1

We will use simple laws of exponents to arrive at our conclusion i.e. the answer.

${( {3}^{ - 1} + {4}^{ - 2} + {6}^{ - 4}) }^{ - 1} \\ \\ = {{( \frac{1}{3} }^{1} + { \frac{1}{4} }^{ 2} + { \frac{1}{6} }^{4} )}^{ - 1} \\ \\ = { \frac{1}{3} }^{ - 1} + { \frac{1}{4} }^{2 \times - 1} + { \frac{1}{6} }^{4 \times - 1} \\ \\ = 3 + 16 + 1296 \\ \\ = 19 + 1296 \\ \\ = 1315$

Remember when a number is raised to the power ( - 1 ) or any negative number then we first take its reciprocal and then simply calculate the required answer.