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The logarithm function
In view of the definition of the logarithm we are going to define the function which assigns the logarithm to every positive number.
Let us plot the graphs of and characterise the following functions:
Let us plot the graphs of and characterise the following functions:
\latex{f: \R \to \R^{+}}, \latex{f(x) = 2^{x}} and \latex{g: \R^{+} \to \R}, \latex{g(x) = \log_2 {x}}.
Since the function \latex{f(x)=2^{x}} is one-to-one and \latex{\log_2 2^{x} = x}, the function \latex{g} is the inverse of the function \latex{f}.
The graph of \latex{g} can be derived from the graph of \latex{f} by reflecting it about the straight line \latex{y = x}. (Figure 22)

\latex{y=2^{x}}
\latex{y=x}
\latex{y=\log_{2}{x}}
Figure 22
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
The function \latex{g} is strictly increasing, its range is the set of real numbers, its zero is \latex{x = 1}. It is a concave function.

Example 1
Let us plot the graphs of and characterise the following functions:
\latex{f:\R\to\R^+}, \latex{f(x)=\left(\frac{1}{3}\right)^{x}} and \latex{g:\R^{+}\to\R}, \latex{g(x)=\log_\frac{1}{3} x }.
Solution
The function \latex{g} and the function \latex{f} are each other's inverses, since \latex{\log_\frac{1}{3}\left(\frac{1}{3}\right)^{x}=x}. The graph of the function \latex{g} can also be derived from the graph of the function \latex{f} by reflecting it about the straight line \latex{y = x}. (Figure 23)
The function \latex{g} is strictly decreasing, its range is the set of real numbers, its zero is \latex{x = 1}. It is a convex function.
In general the function \latex{f: \R^{+} \to \R, f(x) = \log_{a} {x}} is strictly increasing, if \latex{a \gt 1}, and it is strictly decreasing, if \latex{0 \lt a \lt 1}.
The function \latex{f} has a zero at \latex{x = 1}. (Figure 24)
Example 2
Let us give the largest subset of the set of real numbers on which the following functions can be defined, and let us plot the graphs of the functions:
- \latex{f(x) = \log_{2} {x} + 2};
- \latex{g(x) = \log_{2} (x – 2)};
- \latex{h(x) = 3 – \log_{2} (2 – x)}.
Solution (a)
The domain of the function \latex{f(x) = \log_{2} x + 2} is the set of positive numbers.
It takes a value \latex{2} greater at every place of the domain than the function \latex{f_{1}(x) = \log_{2} {x}} does, therefore its graph can be derived from the graph of the function \latex{f_{1}} by translating it by \latex{2} units in the positive direction along the \latex{y}-axis. (Figure 25)
The range of our function is the set of real numbers.
Solution (b)
The domain of the function \latex{g(x) = \log_{2} (x – 2)} is the set of real numbers greater than \latex{2}.
The function \latex{g} takes the same values as the function \latex{f_{1}(x) = \log_{2} {x}} does, only at places \latex{2} greater, therefore its graph can be derived from the graph of the function \latex{f_{1}} by translating it along the \latex{x}-axis by \latex{2} units in the positive direction. (Figure 26)
The zero of the function is: \latex{x = 3}.
Solution (c)
The domain of the function is the set of real numbers less than \latex{2}. From the graph of the function \latex{f_{1}(x) = \log_{2} {x}} the graph of the function \latex{h} can be derived in several steps:
- The graph of \latex{h_{1}(x) = \log_{2} (–x)} can be derived from the graph of \latex{f_{1}} by reflecting it about the \latex{y}-axis.
- Since \latex{\log_{2} {(2 – x)} = \log_{2} [–{(x – 2)]}}, the graph of the function \latex{h_{2}(x) = \log_{2} {(2 – x)}} can be derived from the graph of the function \latex{h_{1}} by translating it along the \latex{x}-axis by \latex{2} units in the positive direction.
- The graph of the function \latex{h_{3}(x) = –\log_{2} {(2 – x)}} can be derived from the graph of \latex{h_{2}} by reflecting it about the \latex{x}-axis.
- The graph of the function \latex{h(x) = 3 – \log_{2} {(2 – x)}} is derived from the graph of \latex{h_{3}} by translating it along the \latex{y}-axis by \latex{3} units in the positive direction. (Figure 27)

Figure 28
\latex{y=\log_{a} x+b\\(a\gt1)}
\latex{y=\log_{a} x\\(a\gt1)}
\latex{y=\log_{a} x+b\\(0\lt a\lt1)}
\latex{y=\log_{a} x\\(0\lt a\lt1)}
\latex{y=\log_{a} x\\(a\gt1)}
\latex{y=\log_{a} (x-c)\\(a\gt1)}
\latex{y=\log_{a} (x-c)\\(0\lt a\lt1)}
\latex{y=\log_{a} x\\(0\lt a\lt1)}
\latex{ 1 }
\latex{ x }
\latex{ b }
\latex{ y }
\latex{ y }
\latex{ y }
\latex{ y }
\latex{ b }
\latex{ 1 }
\latex{ 1 }
\latex{ x }
\latex{ x }
\latex{ x }
\latex{ 1 }
\latex{ c }
\latex{ c+1 }
\latex{ c }
\latex{ c+1 }

Exercises
{{exercise_number}}. The figure shows the height of a plant planted early spring as it changes month by month.
- Fill in the table below.

month
the height of the plant
March
April
May
June
July
August
- Give the \latex{\%} of growth from May till August.
- Give the assignment rule of the function the figure lies on the graph of.

March
May
July
April
June
August
month
\latex{(cm)}
\latex{10 }
\latex{ h }
{{exercise_number}}. Give the largest subset of the set of real numbers on which the following functions can be defined, and plot the graphs of the functions:
- \latex{f(x)=\log_\frac{1}{3} x-3};
- \latex{g(x)=\log_3 x+1};
- \latex{h(x)=\log_\frac{1}{3} \left(x+4\right) };
- \latex{i(x)=\log_5\left(1-x\right) };
- \latex{j(x)=2+\log_\frac{1}{2}x};
- \latex{k(x)=1-\log_\frac{1}{3}\left(x-5\right) };
- \latex{l(x)=2+\frac{1}{5}\times \log_2\left(x-5\right). }
{{exercise_number}}. Based on the graphs give the assignment rules of the logarithm functions below:

\latex{y=f(x)}
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ x }
\latex{ y }
\latex{ 1 }

\latex{y=g(x)}
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ x }
\latex{ 1 }
\latex{ y }

\latex{y=h(x)}
\latex{ 1 }
\latex{ 2 }
\latex{ x }
\latex{ 1 }
\latex{ 2 }
\latex{ y }

\latex{y=i(x)}
\latex{ y }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ x }





