# Where is my mistake in solving the following logartihmic equation

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?

• The value of logarithm can be a negative value. And the formula is correct. – Oldboy Sep 30 '18 at 12:04
• There is a mistake in the step from $\log_{2}{x} - \frac{1}{2}\log_{2}{x} + \frac{1}{4}\log_{2}{x} = 3$ to $\log_{2}{x}(\frac{1}{4} - \frac{1}{2}) = 3$. – mathlove Sep 30 '18 at 12:06
• if $0<x<1$ then $log_ax$ is negative. – Mark Sep 30 '18 at 12:06

$$\log_{2}{x} - \log_{2^2}{x} + \log_{2^4}{x} = 3 \\ \log_{2}{x} - \frac{1}{2}\log_{2}{x} + \frac{1}{4}\log_{2}{x} = 3 \\ (1-\frac{1}{2} + \frac{1}{4})\log_{2}{x} = 3 \\ \frac{3}{4}\log_{2}{x} = 3 \\ \log_{2}{x} = 4 \\ x = 2^4=16$$