Question:

Step 1: Identify the third term in the expansion of $\left(x + x^{\log_{10} x}\right)^5$. The general term in the binomial expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. For the third term, $k=2$ and $n=5$. Let $a=x$ and $b=x^{\log_{10} x}$. $$T_3 = \binom{5}{2} x^{5-2} \left(x^{\log_{10} x}\right)^2$$ $$T_3 = 10 \cdot x^3 \cdot x^{2\log_{10} x}$$ Step 2: Simplify the exponent. Let $y = \log_{10} x$. Then $x = 10^y$. The exponent of $x$ in the third term is $3 + 2\log_{10} x = 3 + 2y$. So, the third term can be written as: $$T_3 = 10 \cdot x^{3+2y}$$ Step 3: Express the third term in terms of $y$. Substitute $x = 10^y$ into the expression for $T_3$: $$T_3 = 10 \cdot (10^y)^{3+2y}$$ $$T_3 = 10 \cdot 10^{y(3+2y)}$$ $$T_3 = 10 \cdot 10^{3y+2y^2}$$ $$T_3 = 10^{1+3y+2y^2}$$ Step 4: Set the third term equal to $10,000,000$ and solve for $y$. Given that the third term is $10,000,000 = 10^7$: $$10^{1+3y+2y^2} = 10^7$$ Equating the exponents: $$1+3y+2y^2 = 7$$ Rearrange into a quadratic equation: $$2y^2+3y-6 = 0$$ Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $$y = \frac{-3 \pm \sqrt{3^2 - 4(2)(-6)}}{2(2)}$$ $$y = \frac{-3 \pm \sqrt{9 + 48}}{4}$$ $$y = \frac{-3 \pm \sqrt{57}}{4}$$ Step 5: Determine the values of $x$. Since $y = \log_{10} x$, we have $x = 10^y$. The two possible values for $y$ are $y_1 = \frac{-3 + \sqrt{57}}{4}$ and $y_2 = \frac{-3 - \sqrt{57}}{4}$. Therefore, the values of $x$ are: $$x_1 = 10^{\left(\frac{-3 + \sqrt{57}}{4}\right)}$$ $$x_2 = 10^{\left(\frac{-3 - \sqrt{57}}{4}\right)}$$