Science:Math Exam Resources/Courses/MATH100/December 2019/Question 12/Solution 1

Solutions of the equation are zeros of the function $f(x)=x-4\log(x)$.

The domain of $f(x)$ is $x>0$.

Since $f'(x)=1-{\frac {4}{x}}$, $f(x)$ is strictly decreasing when $x<4$ and strictly increasing when $x>4$.

This implies that $f(x)$ has a local minimum at $x=4$, and

$f(4)=4-4\log(4)<0$. (This is because $4>e$, so $\log(4)>\log(e)=1$.)

Finally, we have that

$f(1)=1-4\log(1)=1>0$ and $f(16)=16-4\log(16)=16-16\log(2)=16(1-\log(2))>0$.

By the intermediate value theorem, $f(x)$ has at least one zero in the interval $(1,4)$; and at least one zero in the interval $(4,16)$.

But $f(x)$ is strictly decreasing on $(1,4)$; and strictly increasing on $(4,16)$.

Therefore the function $f(x)$ has exactly two zeros: one in the interval $(1,4)$ and another one in the interval $(4,16)$.