JEE Main 2019 — Mathematics Algebra

The general term in the binomial expansion ${(a+b)}^{n}$(a+b)n is ${T}_{r+1}=Crn{a}^{n-r}{b}^{r}.$Tr+1​=Crnan−rbr.

Given, the fourth term in the expansion of ${(\frac{2}{x}+{x}^{{\mathrm{log}}_{8}x})}^{6}$(x2​+xlog8​x)6 is

$${T}_{4}=20\times {8}^{7}$$T4​=20×87

$$\Rightarrow C36{(\frac{2}{x})}^{3}{({x}^{{\mathrm{log}}_{8}x})}^{3}=20\times {8}^{7}$$⇒C36(x2​)3(xlog8​x)3=20×87

$$\Rightarrow \frac{6!}{3!\cdot 3!}{(\frac{2}{x})}^{3}{({x}^{{\mathrm{log}}_{8}x})}^{3}=20\times {({2}^{3})}^{7}$$⇒3!⋅3!6!​(x2​)3(xlog8​x)3=20×(23)7

$$\Rightarrow (\frac{6\times 5\times 4\times 3!}{3\times 2\times 1\cdot 3!}){(\frac{2}{x})}^{3}{({x}^{{\mathrm{log}}_{8}x})}^{3}=20\times {2}^{21}$$⇒(3×2×1⋅3!6×5×4×3!​)(x2​)3(xlog8​x)3=20×221

$$\Rightarrow (20){(\frac{2}{x})}^{3}{({x}^{{\mathrm{log}}_{8}x})}^{3}=20\times {2}^{21}$$⇒(20)(x2​)3(xlog8​x)3=20×221

$$\Rightarrow {[(\frac{2}{x})({x}^{{\mathrm{log}}_{8}x})]}^{3}={({2}^{7})}^{3}$$⇒[(x2​)(xlog8​x)]3=(27)3

$$\Rightarrow (\frac{2}{x})({x}^{{\mathrm{log}}_{8}x})={2}^{7}$$⇒(x2​)(xlog8​x)=27

$$\Rightarrow (\frac{1}{x})({x}^{{\mathrm{log}}_{8}x})={2}^{6}$$⇒(x1​)(xlog8​x)=26

$$\Rightarrow {x}^{{\mathrm{log}}_{8}x-1}={8}^{2}$$⇒xlog8​x−1=82

Taking $\mathrm{log}$log both side to the base $8$8

$$\Rightarrow {\mathrm{log}}_{8}({x}^{{\mathrm{log}}_{8}x-1})={\mathrm{log}}_{8}({8}^{2})$$⇒log8​(xlog8​x−1)=log8​(82)

Using ${\mathrm{log}}_{a}{m}^{n}=n{\mathrm{log}}_{a}m,$loga​mn=nloga​m, we get

$$\Rightarrow ({\mathrm{log}}_{8}x-1)({\mathrm{log}}_{8}x)=2$$⇒(log8​x−1)(log8​x)=2

$\Rightarrow (t-1)(t)=2$⇒(t−1)(t)=2 (Let ${\mathrm{log}}_{8}(x)=t$log8​(x)=t)

$$\Rightarrow {t}^{2}-t-2=0$$⇒t2−t−2=0

$\Rightarrow t=2$⇒t=2 or $-1$−1

$\Rightarrow {\mathrm{log}}_{8}x=2$⇒log8​x=2 or $-1$−1

If ${\mathrm{log}}_{a}x=b,\Rightarrow x={a}^{b},$loga​x=b,⇒x=ab,

$\Rightarrow x={8}^{2}$⇒x=82 or ${8}^{-1}$8−1

$\Rightarrow x=64$⇒x=64 or $\frac{1}{8}.$81​.