Binomial Theorem Question - None

Question

If the fourth term in the binomial expansion of $\left(x^{\sqrt {\left (\frac{1}{1+\log _{10} x}\right)}}+x^{\frac{1}{12}}\right)^{6}$ is equal to 200, and x > 1, then the value of x is

Accepted Answer

Given binomial is $\left(\sqrt{\left(\frac{1}{1+\log _{10} x} ight)}+x^{\frac{1}{12}} ight)^{6}$
Since, the fourth term in the given expansion is 200.
$ herefore^{6} C_{3}\left(x^{\frac{1}{1+\log _{10} x}} ight)^{\frac{3}{2}}\left(x^{\frac{1}{12}} ight)^{3}=200$
$\Rightarrow \quad 20 imes x^{\left[\frac{3}{2\left(1+\log _{10} x ight)}+\frac{1}{4} ight]}=200$
$\Rightarrow \quad x^{\frac{3}{2\left(1+\log _{10} x ight)}+\frac{1}{4}}=10$
$\Rightarrow\left[\frac{3}{2\left(1+\log _{10} x ight)}+\frac{1}{4} ight] \log _{10} x=1$[applying log₁₀ both sides]
$\Rightarrow$[6 + (1 + log₁₀x)] log₁₀x = 4(1 + log₁₀s)
$\Rightarrow$ (7 + log₁₀x) log₁₀x = 4 + 4 log₁₀x
$\Rightarrow$ t² + 7t = 4 + 4t [let log₁₀x = t]
$\Rightarrow$ t² + 3t - 4 = 0
$\Rightarrow$ t = 1, -4 = log₁₀x
$\Rightarrow$ x = 10, 10^-4
Since, x > 1
x = 10

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