The cards below can be arranged to give a proof of the statement

\(\log_c a + \log_c b = \log_c ab\) 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 they would help to make the argument clearer or more convincing.

You might want to print out the cards and rearrange them.

Some blank cards have been included in the cards for printing in case you would like to use them to fill in some details.

One challenge we face when trying to prove something is making it clear what our proof is building on. Our starting point here is that we know how to manipulate indices or powers and we know a relationship between indices and logarithms. We will use results about manipulating indices to prove a result about manipulating logarithms.

Therefore \(ab = c^{x+y}\).

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

Let \(\log_c a =x\) and \(\log_c b =y\).

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

\(c^x = a\) and \(c^y = b\)

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 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 \(\log_c ab\).*