Question:
If the fourth term in the binomial expansion of \(\left(\frac{2}{x}+x^{\log_8 x}\right)^6\) (\(x\gt0\)) is \(20\times8^7\), then a value of \(x\) is :
If the fourth term in the binomial expansion of \(\left(\frac{2}{x}+x^{\log_8 x}\right)^6\) (\(x\gt0\)) is \(20\times8^7\), then a value of \(x\) is :
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Write \(T_4=\binom63\left(\frac2x\right)^3\left(x^{\log_8x}\right)^3\) and put \(t=\log_8x\) to get a quadratic in \(t\).
Updated On: Nov 5, 2026
- \(8^{-2}\)
- \(8^{2}\)
- \(8^{3}\)
- \(8\)
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The Correct Option is B
Solution and Explanation
\(T_4=\binom{6}{3}\left(\frac2x\right)^{3}\left(x^{\log_8x}\right)^{3}=20\cdot\frac{8}{x^3}\cdot x^{3\log_8x}\).
Set \(T_4=20\cdot8^7\): \(x^{3\log_8x-3}=8^6\).
Let \(t=\log_8x\), so \(x=8^t\): \(8^{t(3t-3)}=8^6\Rightarrow 3t^2-3t=6\Rightarrow t^2-t-2=0\Rightarrow t=2\) or \(t=-1\).
So \(x=8^2\) or \(x=8^{-1}\). Only \(8^2\) appears in the options.
Answer: (2) \(8^{2}\)
Set \(T_4=20\cdot8^7\): \(x^{3\log_8x-3}=8^6\).
Let \(t=\log_8x\), so \(x=8^t\): \(8^{t(3t-3)}=8^6\Rightarrow 3t^2-3t=6\Rightarrow t^2-t-2=0\Rightarrow t=2\) or \(t=-1\).
So \(x=8^2\) or \(x=8^{-1}\). Only \(8^2\) appears in the options.
Answer: (2) \(8^{2}\)
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