How does $\log_2(A)-\log_2(B)+\log_2(c)$ not equal $\log_2(\frac{Bc}{A})$

Question

$\log_2(A)-\log_2(B)+\log_2(c)$

How does this equal $\log_2(\frac{Ac}{B})$? Does it not follow from the order of operations that it would be addition first then subtraction?

I'm having hard time understanding why it isn't $\log_2(\frac{Bc}{A})$.

Answer 1

Order of operations says that addition and subtraction are performed left-to-right, so this is \begin{equation*} (\log_2(A)-\log_2(B)) + \log_2(C) = \log_2\left(\frac{A}{B}\right) + \log_2(C) = \log_2\left(\frac{A*C}{B}\right). \end{equation*}

Answer 2

Remember the two fundamental properties of the logarithms: $$\log_xy + \log_xz = \log_xyz$$ and $$ \log_xy^z= z\log_xy. $$

Answer 3

Addition and subtraction have the same precedence and you evaluate them from left to right. Therefore $x-y+z=(x-y)+z$ rather than $x-(y+z)$ since you start on the left and then move right.

This is because subtraction is actually secretly addition, you're just adding the opposite of the second number so $x-y$ is actually $x+(-y)$. So $x-y+z=x+(-y)+z$ and then you do the first addition first. Although actually addition is nice since if you make everything into addition, it doesn't matter what order you do it since addition is associative which means $x+(y+z)=(x+y)+z$ so order doesn't matter.

Since you asked about order of operations I won't mention log rules, but if you have any further questions don't hesitate to ask.