If the third term in the binomial expansion of (1 + xlog2x
)
equals 2560, then a possible value of x is:​

1 thought on “If the third term in the binomial expansion of (1 + xlog2x<br />)<br />equals 2560, then a possible value of x is:​”

1. Appropriate Question

   If the third term in the binomial expansion of

   [tex] \sf \: {\bigg(1 + {x}^{ log_{2}(x) } \bigg) }^{5} \: is \: 2560, \: find \: the \: pssible \: value \: of \: x.[/tex]

   Formula Used :-

   [tex]\rm :\longmapsto\:In \: the \: expansion \: of \: {(x + a)}^{n}, \: the \: general \: term \: is \: [/tex]

   [tex] \red{ \boxed{ \sf{T_{r + 1} \: = \: ^nC_r \: {(x)}^{n – r} \: {a}^{r} }}}[/tex]

   [tex] \blue{ \boxed{ \sf{ ^nC_2\: = \frac{n(n – 1)}{2} }}}[/tex]

   [tex] \red{ \boxed{ \sf{ {x}^{y} = z \: then \: y = log_{x}(z) \: }}}[/tex]

   [tex] \red{ \boxed{ \sf{ {a}^{ log_{a}(x) } \: = \: x}}}[/tex]

   [tex] \red{ \boxed{ \sf{ {a}^{x} \: > \: 0}}}[/tex]

   Let’s solve the problem now!!!

   Given binomial expansion,

   [tex]\rm :\longmapsto\:{\bigg(1 + {x}^{ log_{2}(x) } \bigg) }^{5}[/tex]

   Now,

   It is Given that

   [tex]\rm :\longmapsto\:T_{3} = 2560[/tex]

   [tex]\rm :\longmapsto\:T_{2 + 1} = 2560[/tex]

   [tex]\rm :\longmapsto\:^5C_2 \: {(1)}^{5 – 2}{\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = 2560[/tex]

   [tex]\rm :\longmapsto\:\dfrac{5 \times 4}{2} \: {\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = 2560[/tex]

   [tex]\rm :\longmapsto\:10 \times {\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = 2560[/tex]

   [tex]\rm :\longmapsto\: {\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = 256[/tex]

   [tex]\rm :\longmapsto\: {\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = {(16)}^{2} [/tex]

   [tex]\rm :\longmapsto\: {\bigg({x}^{ log_{2}(x) } \bigg) }= {(16)}[/tex]

   [tex]\rm :\longmapsto\:Taking \: log_{2} \: on \: both \: sides \: we \: get[/tex]

   [tex]\rm :\longmapsto\: log_{2}{\bigg({x}^{ log_{2}(x) } \bigg) }= log_{2} {(16)}[/tex]

   [tex]\rm :\longmapsto\: log_{2}(x) \: log_{2}(x) = log_{2}( {2}^{4} ) [/tex]

   [tex]\rm :\longmapsto\: (log_{2}(x))^{2} = 4[/tex]

   [tex]\rm :\longmapsto\: log_{2}(x) = \pm \: 2[/tex]

   [tex]\rm :\longmapsto\:x = {2}^{ \pm \: 2} [/tex]

   [tex]\rm :\longmapsto\:x = {2}^{\: 2} \: \: \: or \: \: \: x = {2}^{ – 2} [/tex]

   [tex]\bf\implies \:x = 2 \: \: \: or \: \: \: \dfrac{1}{4} [/tex]

   Additional Information :-

   [tex] \red{ \boxed{ \sf{ ^nC_0 \: = \: ^nC_n\: = 1 }}}[/tex]

   [tex] \red{ \boxed{ \sf{ ^nC_1 \: = \: ^nC_{n – 1}\: = n}}}[/tex]

   [tex] \red{ \boxed{ \sf{ ^nC_r\: = \: \frac{n! }{r! \:(n – r) ! } }}}[/tex]

   [tex] \red{ \boxed{ \sf \: \frac{^nC_r}{^{n}C_{r – 1}} \sf{ \: = \: \frac{n – r + 1}{r} }}}[/tex]

   [tex] \red{ \boxed{ \sf{ \: ^nC_r \: + \: ^{n}C_{r – 1} = \: ^{n + 1}C_{r}}}}[/tex]

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