If the third term in the binomial expansion of [( 1 + x^ _2x )^5] equals 2560, then a poss…

Question

If the third term in the binomial expansion
of $${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$ equals 2560, then a possible value of x is -

Accepted Answer

$${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$

$${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$$

$$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$$

$$ \Rightarrow \,\,{x^{2\log 2x}} = 256$$

$$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$$

$$ \Rightarrow 2{({\log _2}x)^2} = 8$$

$$ \Rightarrow \,\,{({\log _2}x)^2} = 4$$

$$ \Rightarrow \,\,{\log _2}x = 2$$  or  - 2

$x = 4$   or  $${1 \over 4}$$

Solutions · 1