If the third term in the binomial expansion of (1+x^_2x)^5 equals 2560, then a possible va…

Question

If the third term in the binomial expansion of $\left(1+x^{\log_2x}\right)^5$ equals 2560, then a possible value of $x$ is:

Accepted Answer

$$T_3=\binom52\left(x^{\log_2x}\right)^2=10\,x^{2\log_2x}=2560\quad\Rightarrow\quad x^{2\log_2x}=256=(\pm16)^2$$

Since $x^{\log_2x}>0$ for $x>0$: $x^{\log_2x}=16$

$$\log_2x\cdot\log_2x=\log_216=4\quad\Rightarrow\quad(\log_2x)^2=4\quad\Rightarrow\quad\log_2x=\pm2\quad\Rightarrow\quad x=4\text{ or }\frac14$$

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