I'm trying to solve the following problem:
$$\log_{2}{x} - \log_{4}{x} + \log_{16}{x} = 3$$
Here is my work:
$$\log_{2}{x} - \log_{2^2}{x} + \log_{2^4}{x} = 3 \\ \log_{2}{x} - \frac{1}{2}\log_{2}{x} + \frac{1}{4}\log_{2}{x} = 3 \\ \log_{2}{x}(\frac{1}{4} - \frac{1}{2}) = 3 \\ \log_{2}{x} \cdot \frac{-1}{4} = 3 \\ \log_{2}{x} = \frac{\frac{3}{1}}{\frac{-1}{4}} \\ \log_{2}{x} = -12$$
The value of logarithm cannot be negative value, so where is my mistake, the formula I'm using is:
$$\log_{a^m}{b^n} = \frac{n}{m} \log_{a}{b}$$ Is this formula correct at all?
$$\log_{2}{x} - \log_{2^2}{x} + \log_{2^4}{x} = 3 \\ \log_{2}{x} - \frac{1}{2}\log_{2}{x} + \frac{1}{4}\log_{2}{x} = 3 \\ (1-\frac{1}{2} + \frac{1}{4})\log_{2}{x} = 3 \\ \frac{3}{4}\log_{2}{x} = 3 \\ \log_{2}{x} = 4 \\ x = 2^4=16$$