+118687 Hi Pyramid :)
You have not specified a base.
Just to make the illustration easier I will make it base 2
Lets look atsome values for $$y=log_2x$$
$$\\log_28=3\qquad $because $8=2^3\qquad (8,3)\\\\
log_24=2\qquad $because $4=2^2\qquad (4,2)\\\\
log_22=1\qquad $because $2=2^1\qquad (2,1)\\\\
log_21=0\qquad $because $1=2^0\qquad (1,0)\\\\
log_2\frac{1}{2}=-1\qquad $because $\frac{1}{2}=2^{-1}\qquad (\frac{1}{2},-1)\\\\
log_2\frac{1}{4}=-2\qquad $because $\frac{1}{4}=2^{-2}\qquad (\frac{1}{4},-2)\\\\
log_2\frac{1}{8}=-4\qquad $because $\frac{1}{8}=2^{-3}\qquad (\frac{1}{8},-3)\\\\$$
Now I can plot these points and then draw the points joining them as accurately as I can.
Now I can estimate other outputs directly from the graph.
Here is the graph
+118687 Hi Pyramid :)
You have not specified a base.
Just to make the illustration easier I will make it base 2
Lets look atsome values for $$y=log_2x$$
$$\\log_28=3\qquad $because $8=2^3\qquad (8,3)\\\\
log_24=2\qquad $because $4=2^2\qquad (4,2)\\\\
log_22=1\qquad $because $2=2^1\qquad (2,1)\\\\
log_21=0\qquad $because $1=2^0\qquad (1,0)\\\\
log_2\frac{1}{2}=-1\qquad $because $\frac{1}{2}=2^{-1}\qquad (\frac{1}{2},-1)\\\\
log_2\frac{1}{4}=-2\qquad $because $\frac{1}{4}=2^{-2}\qquad (\frac{1}{4},-2)\\\\
log_2\frac{1}{8}=-4\qquad $because $\frac{1}{8}=2^{-3}\qquad (\frac{1}{8},-3)\\\\$$
Now I can plot these points and then draw the points joining them as accurately as I can.
Now I can estimate other outputs directly from the graph.
Here is the graph
The basic series expansion for log to the base e (so called natural logarithms) is
log(1 + x) = x - x^2/2 + x^3/3 - x^4/4 + ...
That works providing that x lies within the range minus 1 to plus 1, but excluding minus 1.
There are a number of ways in which the series can be used in order to extend the range, and a change of base formula can be used to produce base 10 tables.
