Logarithmic Equation Involving Trigonometric Functions

Question

Solve the following equation in real numbers: $\log_2(\sin x) + \log_3(\tan x) = \log_4(\cos^2 x) + \log_5(\cot x)$

My approach: $\log_2(\sin x) + \log_3\left(\frac{\sin x}{\cos x}\right) = \log_2(\cos x) + \log_5\left(\frac{\cos x}{\sin x}\right)$

Now the equation can be written as: $\log_2(\sin x) + \log_5(\sin x) + \log_3(\sin x) = \log_2(\cos x) + \log_5(\cos x) + \log_3(\cos x)$

How can I continue my proof?

Answer 1

I think it's better to rearrange like below $$\log_2(\sin x) + \log_3(\tan x) = \log_4(\cos^2 x) + \log_5(\cot x)\\\log_4(\sin^2 x) + \log_3(\tan x) = \log_4(\cos^2 x) + \log_5(\cot x)\\\log_4(\sin^2 x) -\log_4(\cos^2 x) = \log_5(\cot x)- \log_3(\tan x)\\\log_4(\frac {\sin^2 x}{\cos^2 x})=\log_5(\tan x)^{-1}-\log_3(\tan x)\\\log_4(\tan^2x)=-\log_5(\tan x)-\log_3(\tan x)\\$$can you take over?

Second hint:$\boxed{ something \ +=something \ -}$ the only solution is ???

Answer 2

$\log_b(t)=\frac{\ln t}{\ln b}$ hence your equation (when it makes sense, i.e. when $\sin x$ and $\cos x$ are positive) is equivalent to $$\left(\frac1{\ln2}+\frac1{\ln3}+\frac1{\ln5}\right)\ln(\tan x)=0,$$ which you will easily solve.