Alan

This figure should help: the green plane just touches the bottom of the orange function.

If \(\log_9{(x-1)^2}=-1\) then \((x-1)^2=9^{-1}=\frac{1}{9}\)  so  \(x-1=\pm\frac{1}{3}\)

Can you take it from here?

The binomial squared must be in the form  \((\sqrt ax+3)^2\)

When expanded this is \(ax^2+6\sqrt ax+9\)   which means that  \(6\sqrt a=16\)   or  \(a=\frac{64}{9}\)

Various multiples of pi/4 make sin²(x) - 1/2 = 0.   However, none of these result in x being an integer!

If the question is restricted to real numbers, then the argument to the log function must simply be positive.  In this case, log(sin²(x)-1/2) is ok for x = 1 and 2, but is complex for x = 3, so x = 3 would be the smallest positive integer.

My reasoning is as follows:

There are 13^4 possible four letter words in total that can be made with 13 letters.

There are 12^4 possible four letter words that can be made with the 12 letters that aren't A.

Hence there are 13^4 - 12^4 four letter words that contain at least one A.  (13^4 - 12^4 = 7825).

We then need to use similar reasoning for one two and three letter words.

As follows:

Not exactly a derivation, but

It can be proved by induction.

Suppose all ten are lined up randomly in front of you and you have to pin the winning ribbons on the first three - BUT, you've forgotten who they are!

What is the probability that you will just happen to pin the winners ribbon on the overall winner?   1/10

Then, what's the probability that, out of the remaining nine, you will just happen to pin the second place ribbon on the right person?  1/9

Then what's the probability that, out of the remaining eight, you will just happen to pin the third place ribbon on the right person?  1/8

So overall your probability of getting all three right is (1/10)*(1/9)*(1/8) = 1/720

The probability of getting the first digit right in one try is 1/9 (because there are only 9 possibilities, since we can't have a leading zero).

The probability of getting the second digit right, having got the first one right is (1/9)*(1/10) 

The probability of getting the third digit right, having got the first two right is (1/9)*(1/10)² 

...

...

 The probability of getting the seventh digit right, having got the first six right is (1/9)*(1/10)⁶ 

If you have ten independent attempts then the probability is 10*(1/9)*(1/10)⁶  or (1/9)*(1/10)⁵  or 1/900,000

I see it this way: