Solve the inequality 3^x > 500001x.

Step-by-step solution

1. 1

   Turn it into a sign question about one function. Define

   $f(x)=3^x-500001x$

   and look for where $f(x)>0$. This is better than comparing two graphs by eye, because $f$ has structure we can pin down exactly.

2. 2

   Show $f$ is convex, so it has at most two roots.

   $f'(x)=3^x\ln 3-500001,\qquad f''(x)=3^x(\ln 3)^2>0$

   A strictly convex function decreases then increases, crossing zero at most twice. The minimum is where $f'=0$:

   $3^x=\frac{500001}{\ln 3}\approx 455121\quad\Longrightarrow\quad x=\log_3(455121)\approx 11.86$

3. 3

   Handle the left side for free. For $x\le 0$ the left-hand side $3^x$ is positive while $500001x\le 0$, so the inequality holds automatically. At $x=0$ it reads $1>0$ $\checkmark$. So the first root must be slightly to the right of $0$.

4. 4

   Locate the small root. Near $0$, $3^x\approx 1$, so the crossing is close to $x\approx\tfrac{1}{500001}$. Bisecting $f$ on $\left[10^{-12},0.5\right]$ gives

   $x_1\approx 2.0000004\times 10^{-6}$

   and $f$ is positive to the left of it, negative to the right.

5. 5

   Locate the large root by bracketing. $f(12)=531441-6000012<0$ and $f(16)=43046721-8000016>0$, so the second crossing lies in $(12,16)$. Bisection gives

   $x_2\approx 14.3704368$

   Check: $3^{14.3704368}\approx 7.1853\times 10^{6}$ and $500001\times 14.3704368\approx 7.1853\times 10^{6}$, agreeing to six figures $\checkmark$. (A value such as $14.546$ is not a root: there $3^x\approx 8.716\times 10^6$ against $7.273\times 10^6$, a gap of about $1.44\times 10^6$.)

6. 6

   Assemble the solution set. $f>0$ outside the two roots:

   $x<2.0000004\times 10^{-6}\quad\text{or}\quad x>14.3704368$

   Spot-check the middle: at $x=1$, $3<500001$, so the inequality correctly fails $\checkmark$.