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Solve: ${x^{\left( { - 3} \right)}} = 8$. How do I solve for $x$ ?

Answer
VerifiedVerified
483.3k+ views
Hint: First, we shall analyze the given data so that we are able to solve the problem. Here we are given an algebraic equation ${x^{\left( { - 3} \right)}} = 8$and we need to calculate the solution for the given equation. Generally, an algebraic equation is a process of equating some combination of variables and constants to another set of combinations. In this question, we need to simplify the left side of the equation. That is we shall remove the inverse and when we remove the inverse, the $x$term containing the power will get into the denominator and so we can easily find the required answer.
Formula to be used:
The required formula to be applied to this problem is as follows.
${x^{\left( { - a} \right)}} = \dfrac{1}{{{x^a}}}$

Complete step-by-step answer:
The given equation is ${x^{\left( { - 3} \right)}} = 8$and we are asked to solve the equation (i.e. we need to find the solution for the given equation).
Let us consider the given equation ${x^{\left( { - 3} \right)}} = 8$
Now, we shall apply the formula ${x^{\left( { - a} \right)}} = \dfrac{1}{{{x^a}}}$in the above equation.
$ \Rightarrow \dfrac{1}{{{x^3}}} = 8$
We shall take the reciprocal on both sides of the above equation.
$ \Rightarrow {x^3} = \dfrac{1}{8}$
$ \Rightarrow {x^3} = \dfrac{{{1^3}}}{{{2^3}}}$ (Here we have rewritten $8$ as ${2^3}$ and 1 as ${1^3}$ )
$ \Rightarrow {x^3} = {\left( {\dfrac{1}{2}} \right)^3}$
We can note that the powers are the same on both sides of the above equation. So we can compare the base.
Thus, we have $x = \dfrac{1}{2}$ and it is the required solution of the given equation.

Note: First, we need to analyze that the given equation is an algebraic equation. Since we are asked to solve the given equation, we have calculated the appropriate solution of the given equation. Also, there is no need to have a single solution for an equation. An equation can contain more than one solution and even no solution can also exist.