Solve x^2 log2(x) < 100000000.

Step-by-step solution

1. 1

   Fix the domain first. $\log_{2}x$ requires $x>0$, so the whole discussion lives on $(0,\infty)$. Any answer that includes negative $x$ or $x=0$ is wrong before the analysis even starts.

2. 2

   Handle $0<x\le 1$ separately - it costs nothing. On $(0,1)$ we have $\log_{2}x<0$ while $x^{2}>0$, so the product is negative and certainly below $10^{8}$; for example at $x=0.5$ the left side is $-0.25$. At $x=1$ it is exactly $0$. So the entire interval $(0,1]$ satisfies the inequality.

3. 3

   Show the left side is strictly increasing for $x>1$. Differentiating $h(x)=x^{2}\log_{2}x=\dfrac{x^{2}\ln x}{\ln 2}$ gives $h'(x)=\frac{2x\ln x+x}{\ln 2}=\frac{x\left(2\ln x+1\right)}{\ln 2},$ which is positive whenever $\ln x>-\tfrac12$, hence for all $x>1$. A strictly increasing function crosses the level $10^{8}$ exactly once, so the solution set must be a single interval $(0,x_{0})$.

4. 4

   Bracket the crossing point. Evaluating $h$: $h(2900)=96\,730\,450.73,\qquad h(2945)=99\,948\,390.39,\qquad h(2946)=100\,020\,529.48,\qquad h(3000)=103\,956\,721.07.$ The sign of $h(x)-10^{8}$ flips between $2945$ and $2946$, so $x_{0}$ lies there. This is the step to be careful with: an estimate near $2954$ already overshoots the boundary, since $h(2954.5)=100\,634\,817$.

5. 5

   Bisect to the crossing. Refining the bracket $[2945,2946]$ by bisection gives $x_{0}=2945.715457.$ Checking it: $x_{0}^{2}=8\,677\,239.5517$ and $\log_{2}x_{0}=11.524402364$, whose product is $100\,000\,000.0000$ - so $x_{0}$ really is the root of $x^{2}\log_{2}x=10^{8}$.

6. 6

   Assemble the solution set. Combining the two regimes, $0<x<2945.715457,$ i.e. $x\in\left(0,\,2945.715457\right)$ to six decimal places. The endpoint is excluded because the inequality is strict.