Two logarithms using the properties

Statement

Using the properties of logarithms, and without a calculator, find the value of:

• (a) $\log_3\left(\dfrac{1}{2}\right)+\log_3(2)$
• (b) $\log_2 \dfrac{2^{32}}{4^{12}}$

Solution

(a) A sum of logarithms becomes the logarithm of the product: $\log_3\left(\frac{1}{2}\right)+\log_3(2)=\log_3\left(\frac{1}{2}\cdot 2\right)=\log_3 1=0.$

(b) Rewrite everything in base $2$: $4^{12}=\left(2^2\right)^{12}=2^{24}$, hence $\log_2\frac{2^{32}}{2^{24}}=\log_2 2^{32-24}=\log_2 2^{8}=8.$

$\boxed{\text{(a) }0;\qquad \text{(b) }8}$