Solving a Complex Logarithmic Equation with Trigonometric Functions
The given equation is
\[ \log_{64}{ \left( 2^{2x} - \sqrt{3} \cos{x} - 6\sin^2{x} \right) } = x .\]
To solve the equation, let's first observe that the base of the logarithm is 64, which can be written as \(2^6\). So we can make the logarithm base transformation using this fact:
\[ \log_{64}{a} = \frac{\log_{2}{a}}{\log_{2}{64}} = \frac{\log_{2}{a}}{6}. \]
Applying this to the given equation gives us:
\[ \frac{\log_{2}{(2^{2x} - \sqrt{3} \cos{x} - 6\sin^2{x})}}{6} = x .\]
Multiplying both sides by 6 to eliminate the denominator, we get
\[ \log_{2}{(2^{2x} - \sqrt{3} \cos{x} - 6\sin^2{x})} = 6x .\]
Now, let's use the property of logarithms that \(\log_{b}{(b^y)} = y\). Here we have \(2^{2x}\) inside the logarithm, which can be directly extracted:
\[ 2x + \log_{2}{(1 - \sqrt{3} \cdot 2^{-2x} \cos{x} - 6 \cdot 2^{-2x} \sin^2{x})} = 6x .\]
Subtract 2x from both sides:
\[ \log_{2}{(1 - \sqrt{3} \cdot 2^{-2x} \cos{x} - 6 \cdot 2^{-2x} \sin^2{x})} = 4x .\]
To find the values of \(x\) that solve this equation would typically require numerical methods or iterative approximations because of the transcendental nature of the equation, which combines both logarithmic and trigonometric parts.
You may have to use a computational tool or graphing calculator to find the solutions for \(x\). It is not feasible to solve this equation analytically due to its complexity, but numerical methods such as Newton-Raphson or software with equation solver functionality will be useful for finding approximate solutions.
