To find the value of \(x\) in the expression \((x + x^{\log_{10}x})^5\) such that the third term of its expansion is 1,000,000, we will use the Binomial Theorem.
The expression \((x + x^{\log_{10}x})^5\) expands according to the Binomial Theorem:
\((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k}\)
where \(n = 5\), \(x = x\), and \(y = x^{\log_{10}x}\).
The third term (i.e., when \(k=2\)) in the expansion is:
\(\binom{5}{2} x^{5-2} (x^{\log_{10}x})^2\)
This simplifies to:
\(\binom{5}{2} x^{3} \cdot x^{2\log_{10}x} = \binom{5}{2} x^{3 + 2\log_{10}x}\)
Given that this term equals 1,000,000, we write:
\(\binom{5}{2} x^{3 + 2\log_{10}x} = 1000000\)
Calculate \(\binom{5}{2} = 10\).
This gives us:
\(10 x^{3 + 2\log_{10}x} = 1000000 \Rightarrow x^{3 + 2\log_{10}x} = 100000\)
Taking logarithms (base 10) on both sides gives:
\((3 + 2\log_{10}x)\log_{10}x = 5\)
Let \(\log_{10}x = t\). Then, the equation becomes:
\(3t + 2t^2 = 5\)
This is a quadratic equation in \(t\):
\(2t^2 + 3t - 5 = 0\)
Solve this using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 3\), and \(c = -5\).
Calculate the discriminant:
\(b^2 - 4ac = 9 + 40 = 49\)
Thus:
\(t = \frac{-3 \pm \sqrt{49}}{4} = \frac{-3 \pm 7}{4}\)
This gives the solutions:
\(t = 1 \text{ or } t = -\frac{5}{2}\)
These correspond to:
\(\log_{10}x = 1 \Rightarrow x = 10\)
\(\log_{10}x = -\frac{5}{2} \Rightarrow x = 10^{-5/2}\)
Thus, the correct answer is: \(10 \text{ or } 10^{-5/2}\).