Let $a = \log 2, b = \log 3,$ and $c = \log 7$. Find $\log \frac{147}{36}$ in terms of $a, b,$ and $c.$

Question

I know what the function $\log$ means, but I dont' know how to apply to a problem like this:

Let $a = \log 2, b = log 3,$ and $c = \log 7$. Find $\log \frac{147}{36}$ in terms of $a, b,$ and $c.$

Source: Rickards Invitational (Algebra II Individual)

Answer 1

$$\log \frac{147}{36}$$ $$=\log\frac{3\times7^2}{2^2\times 3^2}$$ $$=\log\frac{7^2}{2^2\times 3}$$ $$=2\log 7-2\log 2-\log 3$$ $$=2c-2a-b$$

Hope it helps:)

Answer 2

Hint: $$\frac{147}{36}=\frac{49}{4\cdot 3}=\frac{7^2}{2^2\cdot 3}.$$