JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 9)
If the fourth term in the expansion of $${(x + {x^{{{\log }_2}x}})^7}$$ is 4480, then the value of x where x$$\in$$N is equal to :
3
1
4
2
Explanation
T4 = $${}^7{C_3}{x^4}{({x^{{{\log }_2}x}})^3} = 4480$$
$$ \Rightarrow 35{x^4}{({x^{{{\log }_2}x}})^3} = 4480$$
$$ \Rightarrow {x^4}{({x^{{{\log }_2}x}})^3} = 128$$
take log w.r.t. base 2 we get,
$$4{\log _2}x + 3{\log _2}({x^{{{\log }_2}x}}) = {\log _2}128$$
Let $${\log _2}x = y$$
$$4y + 3{y^2} = 7$$
$$ \Rightarrow y = 1,{{ - 7} \over 3}$$
$$ \Rightarrow {\log _2}x = 1,{{ - 7} \over 3}$$
$$ \Rightarrow $$ $$x = 2,x = {2^{ - 7/3}}$$
$$ \Rightarrow 35{x^4}{({x^{{{\log }_2}x}})^3} = 4480$$
$$ \Rightarrow {x^4}{({x^{{{\log }_2}x}})^3} = 128$$
take log w.r.t. base 2 we get,
$$4{\log _2}x + 3{\log _2}({x^{{{\log }_2}x}}) = {\log _2}128$$
Let $${\log _2}x = y$$
$$4y + 3{y^2} = 7$$
$$ \Rightarrow y = 1,{{ - 7} \over 3}$$
$$ \Rightarrow {\log _2}x = 1,{{ - 7} \over 3}$$
$$ \Rightarrow $$ $$x = 2,x = {2^{ - 7/3}}$$
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