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If the third term in the expansion of $\bigg(\large\frac{1}{x}$$+ x^{log_{10}x}\bigg)^5$ is 1000, then $x$ = ?

$\begin{array}{1 1} 1 \\ 10 \\ 100 \\ 1000 \end{array} $

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1 Answer

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Toolbox:
  • General term in the expansion of $(x+y)^n$ = $T_{r+1}=^nC_rx^{n-r}.y^r$
  • $log_a( b)^n=n.log_a b$
  • $log_a b=\large\frac{1}{log_b a}$
  • If $log_a b=x,$ then $a^x=b$
Given $T_3=1000$
$\Rightarrow^5C_2\big(\large\frac{1}{x}\big)^3$$.\big(x^{log_{10}x}\big)^2=1000$
$\Rightarrow\:10.x^{-3}.\big(\large x^{log_{10}x}\big)^2$$=1000$
$\Rightarrow\:x^{-3}\large x^{2log_{10}x}$$=100$
$\Rightarrow\:\large x^{(2log_{10}x - 3)}$$=100$
$\Rightarrow\:2log_{10}-3=log_x10^2$$=2log_x 10$
$\Rightarrow\:2log_{10}x-3=\large\frac{2}{log_{10}x}$
Let $y=log_{10}x$
$\Rightarrow\:2y-3=\large\frac{2}{y}$
$\Rightarrow\:2y^2-3y-2=0$
$\Rightarrow\:y=2\:\:or\:\:-\large\frac{1}{2}$
Since $y not leq 0, \:y=2$
$\Rightarrow\:log_{10}x=2$ or $x=10^2=100$
answered Sep 23, 2013 by rvidyagovindarajan_1
 

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