Question 54

If $$6^{\sqrt[4]{x}} + 8^{\sqrt[4]{x}} = 10^{\sqrt[4]{x}}$$, then the value of $$x$$ is:

$$6^{\sqrt[\ 4]{x}}+\ 8^{\sqrt[\ 4]{x}}=10^{\sqrt[\ 4]{x}}$$

It is a  pythagoras triplet in the form of $$6^{\ 2}+\ 8^{\ 2}=10^{\ 2}$$

By Comparing , we get 

$$=$$>  $$\sqrt[\ 4]{x}=\ 2$$

$$\ x=\ 2^4$$

$$\ x=\ 16$$

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