Question

The value of $(6^{-1}-7^{-1}{)}^{-1}(5^{-1}-4^{-1}{)}^{-1}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

The correct option is D None of these

Given expression is ${(6^{-1}-7^{-1})}^{-1}{(5^{-1}-4^{-1})}^{-1}$.

From laws of exponents, we know that $a^{-m}={\left(\frac{1}{a}\right)}^m$

Hence, the given expression can be simplified as:

${(\frac{1}{6}-\frac{1}{7})}^{-1}{(\frac{1}{5}-\frac{1}{4})}^{-1}$

Solving the fractions, we get:

$\frac{1}{6}-\frac{1}{7}=\frac{1}{42}$

and

$\frac{1}{5}-\frac{1}{4}=\frac{-1}{20}$

Hence, the expression simplifies to:

${\left(\frac{1}{42}\right)}^{-1}{(-\frac{1}{20})}^{-1}$

Again applying the law of exponent, $a^{-m}={\left(\frac{1}{a}\right)}^m$, we get:

$\left(42{)}^1\right(-20{)}^1$

$\Rightarrow 42\times (-20)=-840$

Hence, ${(6^{-1}-7^{-1})}^{-1}{(5^{-1}-4^{-1})}^{-1}=-840$.