## Solution

$\log_c a + \log_c b = \log_c ab$ for any $a,b > 0$ and $c>0$, but $c \neq 1$.

Can you arrange the cards to give a convincing proof? You may need to include some extra algebraic steps or explanations if you think that would help to make the argument clearer or more convincing.

Here is one possible way of proving the result (with some extra statements added).

We will prove that $\log_c a + \log_c b = \log_c ab$ for any $a,b > 0$ and $c>0$, but $c \neq 1$.

We know how to manipulate powers, so we could try to turn expressions involving logarithms into expressions involving powers. To do this, it might be helpful to name $\log_c a$ and $\log_c b$.

Let $\log_c a =x$ and $\log_c b =y$.

We can rewrite these equations to give us two equations involving powers.

$c^x = a$ and $c^y = b$

We would like to express $\log_c ab$ in terms of $\log_c a$ and $\log_c b$. If we try to express $ab$ in terms of $c$, we might be able to say more about $\log_c ab$.

$ab = c^{x}c^{y}$

Therefore $ab = c^{x+y}$.

Therefore $\log_c ab = \log_c a + \log_c b$, as required.