If the fourth term in the binomial expansion of (1x^1+_10x+x^1/12)^6 is equal to 200, and …

Question

If the fourth term in the binomial expansion of $\left(\dfrac1{x^{1+\log_{10}x}}+x^{1/12}\right)^6$ is equal to 200, and $x>1$, then the value of $x$ is:

Accepted Answer

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

Solutions · 1