Question

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

Answer 1

Appropriate Question

If the third term in the binomial expansion of

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

Formula Used :-

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

$\red{ \boxed{ \sf{T_{r + 1} \: = \: ^nC_r \: {(x)}^{n - r} \: {a}^{r} }}}$

$\blue{ \boxed{ \sf{ ^nC_2\: = \frac{n(n - 1)}{2} }}}$

$\red{ \boxed{ \sf{ {x}^{y} = z \: then \: y = log_{x}(z) \: }}}$

$\red{ \boxed{ \sf{ {a}^{ log_{a}(x) } \: = \: x}}}$

$\red{ \boxed{ \sf{ {a}^{x} \: > \: 0}}}$

Let's solve the problem now!!!

Given binomial expansion,

$\rm :\longmapsto\:{\bigg(1 + {x}^{ log_{2}(x) } \bigg) }^{5}$

Now,

It is Given that

$\rm :\longmapsto\:T_{3} = 2560$

$\rm :\longmapsto\:T_{2 + 1} = 2560$

$\rm :\longmapsto\:^5C_2 \: {(1)}^{5 - 2}{\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = 2560$

$\rm :\longmapsto\:\dfrac{5 \times 4}{2} \: {\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = 2560$

$\rm :\longmapsto\:10 \times {\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = 2560$

$\rm :\longmapsto\: {\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = 256$

$\rm :\longmapsto\: {\bigg({x}^{ log_{2}(x) } \bigg) }^{2} = {(16)}^{2}$

$\rm :\longmapsto\: {\bigg({x}^{ log_{2}(x) } \bigg) }= {(16)}$

$\rm :\longmapsto\:Taking \: log_{2} \: on \: both \: sides \: we \: get$

$\rm :\longmapsto\: log_{2}{\bigg({x}^{ log_{2}(x) } \bigg) }= log_{2} {(16)}$

$\rm :\longmapsto\: log_{2}(x) \: log_{2}(x) = log_{2}( {2}^{4} )$

$\rm :\longmapsto\: (log_{2}(x))^{2} = 4$

$\rm :\longmapsto\: log_{2}(x) = \pm \: 2$

$\rm :\longmapsto\:x = {2}^{ \pm \: 2}$

$\rm :\longmapsto\:x = {2}^{\: 2} \: \: \: or \: \: \: x = {2}^{ - 2}$

$\bf\implies \:x = 2 \: \: \: or \: \: \: \dfrac{1}{4}$

Additional Information :-

$\red{ \boxed{ \sf{ ^nC_0 \: = \: ^nC_n\: = 1 }}}$

$\red{ \boxed{ \sf{ ^nC_1 \: = \: ^nC_{n - 1}\: = n}}}$

$\red{ \boxed{ \sf{ ^nC_r\: = \: \frac{n! }{r! \:(n - r) ! } }}}$

$\red{ \boxed{ \sf \: \frac{^nC_r}{^{n}C_{r - 1}} \sf{ \: = \: \frac{n - r + 1}{r} }}}$

$\red{ \boxed{ \sf{ \: ^nC_r \: + \: ^{n}C_{r - 1} = \: ^{n + 1}C_{r}}}}$