Question

If ${x^{{x^4}}} = 4$, then ${x^{{x^2}}} + {x^{{x^8}}}$ will be equal to what?

Answer 1

To solve this question we are to find a solution of the variable. We will use a hit and trial method to get a solution of the given condition and find a solution of the variable. On finding the value of the variable, we will substitute the value in the second condition and can hence find the required solution.

Complete step by step answer:
The given condition is,
${x^{{x^4}}} = 4$
By hit and trial method, we can see that $\sqrt 2$ is a solution of the condition.
As, ${\sqrt 2 ^{{{\sqrt 2 }^4}}} = {\sqrt 2 ^4} = 4$
Also, $- \sqrt 2$ is a solution of the given condition.
As, $- {\sqrt 2 ^{{{( - \sqrt 2 )}^4}}} = - {\sqrt 2 ^4} = 4$
Therefore, we have,
$x = \sqrt 2$ or $x = - \sqrt 2$,
Therefore, for $x = \sqrt 2$
Substituting these values in ${x^{{x^2}}} + {x^{{x^8}}}$, we get,
${x^{{x^2}}} + {x^{{x^8}}} = {\sqrt 2 ^{{{\sqrt 2 }^2}}} + {\sqrt 2 ^{{{\sqrt 2 }^8}}}$
Now, simplifying the equation, we get,
$\Rightarrow {x^{{x^2}}} + {x^{{x^8}}} = {\sqrt 2 ^2} + {\sqrt 2 ^{16}}$
$\Rightarrow {x^{{x^2}}} + {x^{{x^8}}} = 2 + 256$
$\Rightarrow {x^{{x^2}}} + {x^{{x^8}}} = 258$
And, also, for $x = - \sqrt 2$,
Substituting these values in ${x^{{x^2}}} + {x^{{x^8}}}$, we get,
${x^{{x^2}}} + {x^{{x^8}}} = {( - \sqrt 2 )^{{{( - \sqrt 2 )}^2}}} + {( - \sqrt 2 )^{{{( - \sqrt 2 )}^8}}}$
Now, simplifying the equation, we get,
$\Rightarrow {x^{{x^2}}} + {x^{{x^8}}} = {( - \sqrt 2 )^2} + {( - \sqrt 2 )^{16}}$
$\Rightarrow {x^{{x^2}}} + {x^{{x^8}}} = 2 + 256$
$\therefore {x^{{x^2}}} + {x^{{x^8}}} = 258$

Therefore, we can conclude that the value of ${x^{{x^2}}} + {x^{{x^8}}}$ is $258$.

Note: Now, we have a way that helps us to guess the solution to such conditions or to say equations. We can more easily guess the solutions of such complicated equations and it would be easier and faster for us to solve them. By concerning our guessing, we have to first notice if:
${x^a} = a$
If so, the condition satisfies the above equation, then, it must also be a equivalent equation to the following equation, that is,
${x^{{x^a}}} = {x^a} = a$
If the condition is of such a form that satisfies these conditions, then, in a slightly generalised perspective of our view we can conclude that $x = \sqrt[a]{a}$ is a solution of ${x^{{x^a}}} = {x^a} = a$.