# What is the inflection point of the equation (x^5)(ln(x))?

Let $f \left(x\right) = {x}^{5} \ln \left(x\right)$, then the Product Rule says that $f ' \left(x\right) = 5 {x}^{4} \ln \left(x\right) + {x}^{5} \setminus \cdot \setminus \frac{1}{x} = 5 {x}^{4} \ln \left(x\right) + {x}^{4}$. Now use the Product Rule again to get $f ' ' \left(x\right) = 20 {x}^{3} \ln \left(x\right) + 5 {x}^{4} \setminus \cdot \setminus \frac{1}{x} + 4 {x}^{3} = 20 {x}^{3} \ln \left(x\right) + 9 {x}^{3} = {x}^{3} \left(20 \ln \left(x\right) + 9\right)$.
These equations are only valid for $x > 0$ since $\ln \left(x\right)$ is undefined for $x \setminus \le q 0$. Therefore, $f ' ' \left(x\right) = 0$ when $20 \ln \left(x\right) + 9 = 0$, which means $\setminus \ln \left(x\right) = - \setminus \frac{9}{20}$ and $x = {e}^{- \frac{9}{20}} \setminus \approx 0.6376$. In fact, you can check that $f ' ' \left(x\right)$ changes sign (from negative to positive) as $x$ increases through the value ${e}^{- \frac{9}{20}}$, making $x = {e}^{- \frac{9}{20}}$ the $x$-coordinate of a true inflection point of $f$ (the graph of $f$ will change from concave down to concave up as $x$ increases through this value).