Can You Solve the Equation with Logs: $\log_2 (\cos x)=2\log_3 (\cot x)$?

[anemone]

Solve the equation $\log_2 (\cos x)=2\log_3 (\cot x)$.

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[anemone]

No one answered last week's problem. :(

You can find the proposed solution below:

Spoiler

Before we get started, notice that there are two restrictions set for the given equation:

1. LHS tells us $\cos x>0$ and

2. RHS tells us $\sin x>0$.

This could be turned into a more valuable info if we consider only the domain for $[0,\,2\pi]$, since the function $y=\log_2 (\cos x)$ and $y=2\log_3 (\cot x)$ on both sides are periodic functions with the period of $2\pi$.

This implies we need to only consider the interval $\left(0,\,\dfrac{\pi}{2}\right)$.

We could rewrite the given equation by using the change-of-base formula so that both logarithms have the same base.

$\log_2 (\cos x)=2\log_3 (\cot x)$

$\dfrac{\log_{10} (\cos x)}{\log_{10} 2}=2\dfrac{\log_{10} (\cot x)}{\log_{10} 3}$

$\log_{10} 3\log_{10} (\cos x)=2\log_{10} 2\log_{10} (\cot x)$

$\log_{10} 3\log_{10} (\cos x)=2\log_{10} 2\log_{10} \left(\dfrac{\cos x}{\sin x}\right)$

$\log_{10} 3\log_{10} (\cos x)=2\log_{10} 2\log_{10} \cos x-2\log_{10} 2\log_{10} \sin x$

$2\log_{10} 2\log_{10} \sin x=2\log_{10} 2\log_{10} \cos x-\log_{10} 3\log_{10} (\cos x)$

$2\log_{10} 2\log_{10} \sin x=\log_{10} \cos x(2\log_{10} 2-\log_{10} 3)$

Notice that the function on the LHS is continuous increasing while the function on the RHS is continuous decreasing over the interval $\left(0,\,\dfrac{\pi}{2}\right)$.

We can conclude based on that observation that there exists only one intersection point for the both functions in that interval, and it's not hard to see $x=\dfrac{\pi}{3}$ is the answer.

The solution for this problem is hence $x=\dfrac{\pi}{3}+2k\pi$, where $k\in\mathbb{Z}$.