Question

Solve the given using identities: ${{\left( -3 \right)}^{-2}}=\dfrac{1}{k}$ .

Answer 1

Hint: The question is related to the understanding of the concepts related to exponents and powers. Start by using the identity that ${{a}^{-b}}=\dfrac{1}{{{a}^{b}}}$ . After that cross multiply and further simplify to get the value of k. Remember that the square of a negative number is always positive.

Complete step-by-step answer:
The equation given in the question is:
${{\left( -3 \right)}^{-2}}=\dfrac{1}{k}$
Looking at the equation, we can say that the value of k can never be zero.
Now to simplify the equation, we will use the identity ${{a}^{-b}}=\dfrac{1}{{{a}^{b}}}$ . On doing so, we get
$\dfrac{1}{{{\left( -3 \right)}^{2}}}=\dfrac{1}{k}$
Now for converting the equation to linear which will make it easier to solve, we will cross multiply. On cross-multiplication, we get
 $k={{\left( -3 \right)}^{2}}$
Now, we know that the square of a real negative number is always a positive real number. So, the square of -3 is equal to 9. If we use this in our equation, the value of k comes out to be:
 $k=9$
Hence, we can conclude that the value of k that satisfies the equation ${{\left( -3 \right)}^{-2}}=\dfrac{1}{k}$ is equal to 9.

Note: It is always a good practice to verify the solution by putting the final values of variables that you get in the parent equation and confirming whether the equation is satisfied or not. You should also be very careful about the domain of the equation, as in the above equation k must not be equal to 0 because this would make the denominator of the right-hand side of the equation 0 making the right-hand side undefined.