
If \[{x^{{x^4}}} = 4\], then ${x^{{x^2}}} + {x^{{x^8}}}$ will be equal to what?
Answer
493.5k+ views
Hint: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 ${x^{{x^a}}} = {x^a} = a$.
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 ${x^{{x^a}}} = {x^a} = a$.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

