Alan

ln(a) = b  means that a = e^b, so ln(x+9) = 1 means x+9 = e¹ so x = e - 9 or in approximate decimal form:

$${\mathtt{x}} = {\mathtt{e}}{\mathtt{\,-\,}}{\mathtt{9}} \Rightarrow {\mathtt{x}} = -{\mathtt{6.281\: \!718\: \!171\: \!540\: \!954\: \!8}}$$

A logistic curve (function f(x) below) seems to fit reasonably well.

However, there is no such thing as the equation, in general, when creating curves that are a best-fit to a few data points.  You can fit lots of curves to this data; some will fit better than others.  Did you have a particular type of curve in mind?  Your title suggests you want a logarithmic fit of some sort, but straight line fits to ln(y) vs x; y vs ln(x) or ln(y) vs ln(x) all give very poor fits! 

With real numbers you cannot have the logarithm of a negative number.

With complex numbers

$$\log_2{(-m)}= \log_2{(m)}+\frac{(2n+1)\pi}{\ln2}*i$$ 

where m is positive and n is any (positive or negative) integer.

$$\frac{2}{5}\times\frac{3}{7}=\frac{2\times3}{5\times7}=\frac{6}{35}$$

In approximate decimal representation:

$${\frac{{\mathtt{6}}}{{\mathtt{35}}}} = {\mathtt{0.171\: \!428\: \!571\: \!428\: \!571\: \!4}}$$

1. Try adding up, say, 127 three hundred and eighty nine times.

2. Try adding up 8 one seventh times. 

3. Try calculating 7! (that's 7 factorial = 7*6*5*4*3*2*1) using repeated addition.

Try these and you'll quickly come to appreciate multiplication!

Just the one!

If t_n = n-1 then t₉₉ = 98,  t₁₀₀ = 99 (= t₉₉ + 1) and t₁₁ = 10 so that t₁₁² - 1 = 99 (= t₁₀₀)

There are an infinite number of functions that would satisfy this!  I think the question is incomplete.

However, the process is more important than the result here, so, although you might not get full marks you should get some for showing the correct way of doing it!