Fourth term (T₄)

= $${}^6{C_3}{\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} } \right)^3}{\left( {{x^{{1 \over {12}}}}} \right)^3}$$

= $$20{\left( {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} \right)^{{3 \over 2}}}\left( {{x^{{1 \over 4}}}} \right)$$

= $$20 \times {x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right){3 \over 2}}} \times {x^{{1 \over 4}}}$$

= $$20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$$

Given, T₄ = 200

$$ \therefore $$ $$20 \times {x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$$ = 200

$$ \Rightarrow $$ $${x^{\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right)}}$$ = 10

Taking log₁₀ on both sides

$$\left( {{3 \over {2\left( {1 + {{\log }_{10}}x} \right)}} + {1 \over 4}} \right){\log _{10}}x$$ = 1

put log₁₀ x = t

$$\left( {{3 \over {2\left( {1 + t} \right)}} + {1 \over 4}} \right)t$$ = 1

$$ \Rightarrow $$ $$\left( {{{\left( {1 + t} \right) + 6} \over {4\left( {1 + t} \right)}}} \right) \times t$$ = 1

$$ \Rightarrow $$ t² + 7t = 4 + 4t

$$ \Rightarrow $$ t² + 3t - 4 = 0

$$ \Rightarrow $$ (t + 4)(t - 1) = 0

$$ \Rightarrow $$ t = 1 or t = - 4

$$ \therefore $$ log₁₀ x = 1

$$ \Rightarrow $$ x = 10

or log₁₀ x = - 4

$$ \Rightarrow $$ x = 10^-4

But as x > 1 so x $$ \ne $$ 10^-4

$$ \therefore $$ x = 10