Logarithms of Quotients

We have already seen that the logarithm of a product is the sum of the logarithms of the factors:

$\displaystyle \log_b(xy) = \log_bx + \log_by$

Similarly, the logarithm of the ratio of two quantities is the difference of the logarithms:

$\displaystyle \log_b\left( \frac{x}{y}\right) = \log_bx - log_by.$ 

We can show that this is true by the following example:

Let $u=\log_b x$ and $v=\log_b y$.

Then $b^u=x$ and $b^v=y.$ 

Then: 

$\displaystyle \begin{aligned} \log_b\left(\frac{x}{y}\right)&=\log_b\left({b^u \over b^v}\right)\\ &= \log_b(b^{u-v}) \\ &=u-v\\ &= \log_b x - \log_b y \end{aligned}$

Another way to show that this rule is true, is to apply both the power and product rules and the fact that dividing by $y$ is the same is multiplying by $y^{-1}.$ So we can write:

 $\displaystyle \begin{aligned} \log_b\left(\frac{x}{y}\right)&=\log_b(x\cdot y^{-1})\\ & = \log_bx + \log_b(y^{-1})\\& = \log_bx -\log_by \end{aligned}$

Example: write the expression $\log_2\left({x^4y^9 \over z^{100}}\right)$ in a simpler way 

By applying the product, power, and quotient rules, you could write this expression as: 

$\displaystyle \log_2(x^4)+\log_2(y^9)-\log_2(z^{100}) = 4\log_2x+9\log_2y-100\log_2z.$