Solution

One of the key differences between solving exponential equations and logarithmic equations, is that the logarithmic equations will have domain restrictions. In general, if \(\log(A)\) is in your equation, you must ensure that \(A\gt 0\). That is to say, the inside of any logarithm must always work out to be strictly positive.

For us, that means \[ \solve{ 12-x&\gt& 0\\ 12&\gt&x\\ x&\lt &12 } \] So, in solving the equation, we must reject any answer(s) we find that result in \(x\) being 12 or greater. Let's now solve the equation and see what we get. If you are wondering how, we will rely on the definition of logarithm to convert this logarithmic equation into an exponential one: \[ \solve{ \log_7(12-x)&=&2\\ 7^2&=&12-x\\ 49&=&12-x\\ 37&=&-x\\ -37&=&x } \] After checking that this fits within the domain, we can then accept our answer is \(-37\).

As a reminder, we can use our graphing utilities to assist us, especially when it comes to checking our work. Here is a graph of both sides of the equation and the corresponding intersection: Note that the domain of the function is exactly as we had previously calculated. To get the graph window, I used our solution combined with the domain restriction.