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The exponential and the logarithm function
Let us summarise the things we learnt about the exponential and the logarithm function.
If \latex{ a\gt0 }, then the function \latex{x\mapsto a^x, x\in \R}:
- is strictly increasing in the case \latex{ a\gt1 },
- is strictly decreasing in the case \latex{ 0\lt a\lt1 },
- is constant in the case \latex{ a = 1 }: its value is \latex{ 1 }(Figure 1).
\latex{ y }
\latex{ x }
\latex{ 1 }
\latex{ 1 }
\latex{ 1^x }
Figure 1
The value of the function \latex{x\mapsto a^x} is positive for every \latex{x\in \R}, and its graph intersects the \latex{ y }-axis at the point (\latex{ 0; 1 }).
If \latex{a\gt 0,\;a\neq 1}, then the function \latex{x\mapsto \log _ax,\;x\in \R^+} (the inverse of the function \latex{x\mapsto a^x}):
- is strictly increasing in the case of \latex{ a\gt1 },
- is strictly decreasing in the case of \latex{ 0\lt a\lt1 } (Figure 2),
- if \latex{ a\gt1 }, then the value of the function \latex{x\mapsto \log _ax}
is negative in the case \latex{ 0\lt x\lt1 },
is positive in the case \latex{ 1\lt x },
is \latex{ 0 } in the case \latex{ x = 1 };
- if \latex{ 0\lt a\lt1 }, then the value of the function \latex{x\mapsto \log _ax}
is positive in the case \latex{ 0\lt x\lt1 },
is negative in the case \latex{ 1\lt x }
is \latex{ 0 } in the case \latex{ x = 1 }.
\latex{y=\log_ax}
\latex{a\gt1}
\latex{a\gt1}
\latex{y=\log_ax}
\latex{0\lt a\lt1}
\latex{0\lt a\lt1}
\latex{ 1 }
\latex{ 1 }
\latex{ y }
\latex{ x }
Figure 2
Example 1
Let us plot the graphs of the following functions:
- \latex{x\mapsto 2^{-x},x\in \R };
- \latex{x\mapsto \log _2\left(2x\right),x\gt 0}.
Solution (a)
Let us use the powers identities:
\latex{2^{-x}=\frac{1}{2^x}=\left\lgroup\frac{1}{2} \right\rgroup^x}.
The image of the function \latex{x\mapsto \left\lgroup\frac{1}{2} \right\rgroup^x} is known, it is strictly decreasing, the value of the function is \latex{ 1 } at the place \latex{ x = 0 }. (Figure 3)
\latex{x= \left\lgroup\frac{1}{2} \right\rgroup^x}
\latex{ 1 }
\latex{ 1 }
\latex{ y }
\latex{ x }
Figure 3
Solution (b)
Now we use the logarithmic identities:
\latex{\log _2\left(2x\right) =\log_22+\log_2x=\log _2x+1}.
The graph of the function \latex{x\mapsto\log_2x+1} is derived from the graph of the function \latex{x\mapsto\log_2x} by translating it by \latex{ 1 } unit in the positive direction of the \latex{ y }-axis. (Figure 4)
\latex{y=\log _2x+1}
\latex{ 1 }
\latex{ \frac {1}{2} }
\latex{ x }
\latex{ 1 }
\latex{ y }
Figure 4
Example 2
Let us plot the graphs of the following functions:
- \latex{x\mapsto 2^{\left|x\right| },x\in \R};
- \latex{x\mapsto \log _2\left(-x\right),x\lt 0};
- \latex{x\mapsto \log _2\left|x\right| ,x\in \R,x\neq 0};
- \latex{x\mapsto\big|\log_2{\lvert x\rvert}\big|,x\in\R,x\neq0}.
Solution (a)
Let us use the definition of the absolute value and the powers identities:
\latex{2^{\lvert x\rvert}=\begin{cases}2^x,\;\text{if}\;\;x\geq0;\\2^{-x}=\frac{1}{2^x}=\left(\frac{1}{2}\right)^x,\;\text{if}\;\;x\lt0.\end{cases}}
According to this the graph of the function can be seen in Figure 5.
\latex{y=2^{\left|x\right| }}
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
Figure 5
Solution (b)
The graph of the function is derived from the graph of the function \latex{x\mapsto\log_2x} by reflecting it about the \latex{ y }-axis (Figure 6).
\latex{y=\log_2x}
\latex{y=\log_2(-x)}
\latex{ 1 }
\latex{ -1 }
\latex{ x }
\latex{ y }
Figure 6
Solution (c)
Based on the definition of the absolute value the expression defining the function can be written as follows:
\latex{\log{2}\lvert x\rvert=\begin{cases}\log_2x,\;\text{if}\;\;x\gt0;\\\log_2(-x),\;\text{if}\;\;x\lt0.\end{cases}}
Based on this the graph of the function can be seen in Figure 7. The graph is symmetric about the \latex{ y }-axis; the function is even.
Solution (d)
The graph of the function is derived from the graph represented in part (c) by reflecting the parts of the graph under the \latex{ x }-axis about the \latex{ x }-axis (Figure 8).
\latex{y=\log _2\left|x\right|}
\latex{-1}
\latex{y}
\latex{x}
\latex{1}
Figure 7
\latex{-1}
\latex{y=\big|\log_2{\lvert x\rvert}\big|}
\latex{y}
\latex{x}
\latex{1}
Figure 8
We say that the function \latex{ f } is convex (concave) on an interval \latex{ I } of its domain, if for an arbitrary element \latex{ a\lt b } of \latex{ I } the graph of the function \latex{ f } is below (above) the straight line lying on the points \latex{ A(a, f(a)) } and \latex{ B(b, f(b)) } between \latex{ a } and \latex{ b }. (Figure 9)
convex
concave
\latex{ A }
\latex{ B }
\latex{ b }
\latex{ a }
\latex{ x }
\latex{ f }(\latex{ a })
\latex{ f }(\latex{ b })
\latex{ f }(\latex{ b })
\latex{ f }(\latex{ a })
\latex{ B }
\latex{ A }
\latex{ A }
\latex{ F }
\latex{ B }
\latex{ P }
\latex{ a }
\latex{ a }
\latex{ b }
\latex{ b }
\latex{ x }
\latex{ x }
\latex{ y }
\latex{ y }
\latex{ y }
\latex{ \frac {a+b}{2} }
Figure 9
It can be verified that if the function \latex{ f } is continuous on the interval \latex{ I }, and for any \latex{ a\lt b } element of \latex{ I } the midpoint of the line segment \latex{ AB } is above (below) the graph of \latex{ f }, then \latex{ f } is convex (concave) on \latex{ I }.
Example 3
Let us verify that the function \latex{x\mapsto a^x\left(a\gt 0,\;a\neq 1\right)} is convex, i.e. if \latex{ x1\lt x2 }, then \latex{a^{\frac{x_1+x_2}{2}}\lt \frac{a^{x_1}+a^{x_2}}{2}} (Figure 10).
Solution
The case \latex{ a\gt 1 } can be seen in Figure 10, the case \latex{ 0\lt a\lt1 } is very similar.
The length of the line segment \latex{ QT } is \latex{\frac{a^{x_1}+a^{x_2}}{2}}, since \latex{ Q } is the midpoint of the line segment \latex{ AB }.
The length of the line segment \latex{ PT } is \latex{a^{\frac{x_1+x_2}{2}}}, which can be written as follows according to the identities of powers: \latex{\sqrt{a^{x_1} \times a^{x_2}}}.
\latex{\frac{x_1+x_2}{2}}
\latex{x_1}
\latex{x_2}
\latex{a^{x_2}}
\latex{a^{x_1}}
\latex{y=a^x}
\latex{ Q }
\latex{ A }
\latex{ P }
\latex{ B }
\latex{ T }
\latex{ x }
\latex{ y }
Figure 10
So we have to verify that
\latex{\sqrt{a^{x_1}\times a^{x_2}} \lt \frac{a^{x_1}+a^{x_2}}{2}}.
This inequality holds because of the inequality of arithmetic and geometric means, because \latex{a^{x_1},a^{x_2}\gt0} and \latex{a^{x_1}\neq a^{x_2}}.
Example 4
Let us verify that the function \latex{x \mapsto \log_a x}, \latex{a \gt1}, \latex{x\in\R^+} is concave on the interval \latex{]0; +\infty[} (Figure 11), i.e.
if \latex{0\lt x_1\lt x_2}, then \latex{\log_a\frac{x_1+x_2}{2}\gt\frac{\log_ax_1+\log_ax_2}{2}.\;\;\;\;\;\;\;\;\;\;\tag 1}
Solution
By using the logarithmic identities the right-hand-side of the inequality can be written as follows:
\latex{\frac{\log_ax_1+\log _ax_2}{2} =\log_a\sqrt{x_1 \times x_2}}.
So it is enough to verify that
\latex{\log _a\frac{x_1+x_2}{2}\gt \log_a\sqrt{x_1 \times x_2}}.
Because of the inequality of arithmetic and geometric means
\latex{\frac{x_1+x_2}{2}\gt \sqrt{x_1 \times x_2}}.
\latex{\frac{x_1+x_2}{2}}
\latex{x_2}
\latex{x_1}
\latex{y=\log_ax}
\latex{ Q }
\latex{ B }
\latex{ A }
\latex{ P }
\latex{ x }
\latex{ y }
\latex{ T }
Figure 11
as \latex{x_1,x_2\gt 0,\;x_1\neq x_2,} and in the case of \latex{ a\gt1 } the logarithm function with base a is strictly increasing on its domain. It implies that the inequality to be proven is true.
If \latex{ 0\lt a\lt1 }, then the function \latex{x\mapsto \log _ax} is convex on the interval \latex{\left]0;+\infty \right[ }. It can be proven similarly. In this case the logarithm function with base a is strictly decreasing, so the relation of the inequality (\latex{ 1 }) holds exactly reversed.
Exercises
{{exercise_number}}. Plot the graphs of the following functions:
- \latex{x\mapsto 2^{2x},x\in \R};
- \latex{x\mapsto \log _2x^2,\;x\neq 0}.
{{exercise_number}}. Plot the graphs of the following functions:
- \latex{x\mapsto \left|\log _2x\right|,x\in \R,x\gt 0 };
- \latex{x\mapsto 2^{\log_x},x\in \R,x\gt 0};
- \latex{x\mapsto \log_22^x,x\in \R}.
{{exercise_number}}. Plot the graphs of the following two functions respectively, and compare them:
- \latex{x\mapsto \log _2x^2,x\neq 0}; and \latex{x\mapsto 2\times \log _2x,\;x\gt 0};
- \latex{x\mapsto \log _2\sqrt{x^2} ,x\neq 0} and \latex{x\mapsto \log _2x,\;x\gt 0}.
{{exercise_number}}. Plot the graphs of the following functions, and characterise the functions (where are they increasing or decreasing, do they have an extreme value etc):
- \latex{x\mapsto \log _2\left(x-2\right),x\gt 2 };
- \latex{x\mapsto \left|\log _\frac{1}{2} \left(x+1\right) \right|,x\gt -1 };
- \latex{x\mapsto \left\lgroup\frac{1}{3} \right\rgroup^{\left|x+2\right| },x\in \R };
- \latex{x\mapsto 2^{\left|x+1\right| },x\in \R }.