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Power with irrational exponent, exponential function
Let us consider the function \latex{f: \Z \to \R}, \latex{f(x) = 2^{x}}.
Some of the points of the function can be plotted. (Figure 6)
The graph also shows and it can also be checked by calculation that the function \latex{f} is strictly increasing.
Some of the points of the function can be plotted. (Figure 6)
The graph also shows and it can also be checked by calculation that the function \latex{f} is strictly increasing.
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In view of the powers with fractional exponent the following function can be defined: \latex{ g: ℚ \to\R}, \latex{g(x) = 2^x}.
Is the function \latex{g} also strictly increasing?
Is the function \latex{g} also strictly increasing?
Figure 6
\latex{y=2^{x}}
\latex{(x \in \Z )}
\latex{(x \in \Z )}
\latex{ y }
\latex{ 1 }
\latex{ 1 }
\latex{ x }
Let \latex{\frac{p}{q}\lt\frac{k}{l}}, where \latex{k, l, p, q \in \Z} and \latex{q, l \geq 2}. Thus the condition means that \latex{p \times l\lt k\times q}.
\latex{2^{\frac{p}{q}}=\sqrt[q]{2^{p}}=\sqrt[q\times l]{2^{p \times l}} \lt \sqrt[q\times l]{2^{k \times q}} =\sqrt[l]{2^k}=2^{\frac{k}{l}}}.
\latex{\huge\vert}
definition
definition
\latex{\huge\vert}
identity E)
identity E)
\latex{\huge\vert}
\latex{p \times l \lt k \times q}
and these are integers
\latex{p \times l \lt k \times q}
and these are integers
\latex{\huge\vert}
identity E)
identity E)
\latex{\huge\vert}
definition
definition
We get that the function \latex{g} is also strictly increasing.
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The following question arises: can \latex{2^{x}} be defined if \latex{x} is an irrational number.
For example how can we define \latex{2^{\sqrt{5}}} ?
It is practical to define it so that the function \latex{h:\R\to\R, h(x) = 2^x} will also be strictly increasing.
For example how can we define \latex{2^{\sqrt{5}}} ?
It is practical to define it so that the function \latex{h:\R\to\R, h(x) = 2^x} will also be strictly increasing.
\latex{2\lt \sqrt{5} \lt 3} thus \latex{2^2\lt 2^{\sqrt{5}} \lt 2^3};
\latex{2.2\lt \sqrt5 \lt2.3} thus \latex{2^{2.2} \lt 2^{\sqrt5} \lt 2^{2.3}};
\latex{2.23\lt \sqrt5 \lt2.24} thus \latex{2^{2.23} \lt 2^{\sqrt5} \lt 2^{2.24}};
\latex{2.236\lt \sqrt5 \lt2.237} thus \latex{2^{2.236} \lt 2^{\sqrt5} \lt 2^{2.237}};
\latex{\vdots}
It can be proven that there is only one real number which satisfies the condition regarding strict increase; let it be the value of \latex{2^{\sqrt{5}}}.
\latex{(2^{\sqrt{5}}\approx 4.51111}.)
We can similarly define \latex{2^{x}} to an arbitrary irrational exponent.
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Every power \latex{a^{x}} can be defined with the same process in the case of \latex{x \in\R}, if \latex{a \gt 1}.
If \latex{0 \lt a \lt 1}, then the values of \latex{a^{x}} can be obtained from the strict decrease of the corresponding function for every real \latex{x}.
If \latex{0 \lt a \lt 1}, then the values of \latex{a^{x}} can be obtained from the strict decrease of the corresponding function for every real \latex{x}.
If \latex{a = 1}, then let \latex{1^{x} = 1} for every \latex{x \in\R}.
It can be proven that the identities of powers stay true for the power with irrational exponent that can only be defined this way.
It can be proven that the identities of powers stay true for the power with irrational exponent that can only be defined this way.
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DEFINITION: The function \latex{f: \R\to \R}, \latex{f(x) = a^{x}}; \latex{a \gt 0} is called the exponential function. (Figure 7)
\latex{y=a^{x}} \latex{\left(0\lt a\lt1\right)}
\latex{y=a^{x}} \latex{\left(a\gt1\right)}
\latex{y=1^{x}}
Figure 7
\latex{ 1 }
\latex{ 1 }
\latex{ x }
\latex{ y }
\latex{ y }
\latex{ y }
\latex{ x }
\latex{ x }
\latex{ 1 }
\latex{ 1 }
\latex{ 1 }
\latex{ 1 }
It can be proven that if \latex{a \neq 1}, then the range of the exponential functions is the set of positive real numbers, and they take every positive value once and only once.
Example 1
Let us plot the graphs of the following functions:
\latex{f:\R \to \R}, \latex{f(x)=2^{x}}; \latex{g:\R \to \R}, \latex{g(x)=3^{x}}; \latex{h:\R \to \R}, \latex{h(x)=10^{x}}.
Figure 8
\latex{y=10^{x}}
\latex{y=3^{x}}
\latex{y=2^{x}}
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
Solution
Since \latex{a^{0} = 1} if \latex{a \gt 0}, the point \latex{(0; 1)} is a point of the graph of every exponential function. (Figure 8)
Example 2
Let us plot the graphs of the following functions:
- \latex{f:\R \to \R}, \latex{f(x)=2^{x}} and \latex{g:\R \to \R}, \latex{g(x)=\left(\frac{1}{2}\right)^x};
- \latex{f:\R \to \R}, \latex{f(x)=3^{x}} and \latex{g:\R \to \R}, \latex{g(x)=\left(\frac{1}{3}\right)^x}.
\latex{y=\left(\frac{1}{2}\right)^x}
\latex{y=2^{x}}
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
Solution
The graphs of the exponential functions with bases \latex{a} and \latex{\frac{1}{a}} are symmetric about the \latex{y}-axis, since \latex{\left(\frac{1}{a}\right)^{x}}= \latex{\left(a^{-1}\right)^{x}=a^{-x}}. (Figure 9)
\latex{y=\left(\frac{1}{3}\right)^x}
\latex{y=3^{x}}
Figure 9
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
Example 3
Let us plot the graphs of the following functions:
- \latex{f:\R \to \R}, \latex{f(x)=2^{x}-3};
- \latex{g:\R \to \R}, \latex{g(x)=2^{x-3}}.
Solution (a)
At every place of the domain the function takes a value \latex{3} less than the function \latex{2^{x}} does, so the graph of the latter one should be translated by \latex{3} units to the negative direction (downwards) along the \latex{y}-axis. (Figure 10)
The function is strictly increasing; its range is the set of numbers greater than \latex{–3}; its zero is between \latex{1} and \latex{2}, the exact value cannot be determined based on our current knowledge.
Solution (b)
The function takes the same values as the function \latex{2^{x}} does, only at places \latex{3} greater, so its graph can be derived by translating the graph of \latex{2^{x}} by \latex{3} units along the \latex{x}-axis to the positive direction (to the right). (Figure 11)
Figure 10
\latex{y=2^{x}}
\latex{y=2^{x}-3}
\latex{ 1 }
\latex{ y }
\latex{ 1 }
\latex{ x }
\latex{ -3 }
\latex{y=2^{x-3}}
\latex{y=2^{x}}
Figure 11
\latex{ 1 }
\latex{ y }
\latex{ 1 }
\latex{ 3 }
\latex{ x }
Example 4
Let us plot the graph of the function \latex{h: \R \to \R}, \latex{h(x) =\left(\frac{1}{2}\right)^{x+3} }.
Solution
The graph of the function can be derived from the graph of the function \latex{\left(\frac{1}{2}\right)^{x} } by translating it by \latex{3} units along the \latex{x}-axis to the negative direction. (Figure 12)
The function will be strictly decreasing; its range will be the set of positive numbers.
\latex{y=\left(\frac{1}{2}\right)^x}
\latex{y=\left(\frac{1}{2}\right)^{x+3}}
Figure 12
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
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The processes of nature can often be described with the help of the exponential function with the base \latex{e} well. \latex{e} is an irrational number, its approximate value is \latex{2.7183}.
Example 5
The air pressure in the atmosphere of Earth can be given with the following expression with a good approximation as the function of height:
\latex{p(h) = p_{0} \times e^{–0.125 \times h}},
where \latex{p_{0} = 10^{5}} Pa is he air pressure measured at sea level and \latex{h} is the height measured in kilometres.
- What is the air pressure at a height of \latex{5.5\,km}?
- What is the percentage of the decrease of the air pressure pressing our eardrum if we go from Szeged (located in Hungary, at \latex{84} m above sea level) to Kékestetõ (located in Hungary, at \latex{1,014\,m} above sea level)?
Solution (a)
\latex{p(5.5)=10^{5}\times e^{-0.125\times 5.5}=10^{5}\times e^{-0.6875}=\\=10^{5}\times 0.5028= 5.028\times 10^{4} \text{Pa}.}
The power of \latex{e} can be calculated with the help of a pocket calculator.
The result shows that the air pressure measured at a height of \latex{5.5\,km} is about half of the air pressure measured at the sea level.
Solution (b)
Let us calculate the quotient of the air pressures measured at the two given places.
\latex{\frac{\text{p(Kékes)}}{\text{p(Szeged)}}=\frac{p_{0}\times e^{-0.125\times 1.014}}{p_{0}\times e^{-0.125\times 0.084}}=e^{-0.125\times 0.93}=\\=e^{-0.11625}=0.89025.}
So the air pressure pressing our eardrum decreases by about \latex{11\%} if we go from Szeged up to Kékestetõ.
Exercises
{{exercise_number}}. Plot the graphs of and characterise the following functions:
- \latex{f:\R \to \R}, \latex{f(x)=2^{x}+1};
- \latex{g:\R \to \R}, \latex{g(x)=2^{x+2}};
- \latex{h:\R \to \R}, \latex{h(x)=2^{x-2}-2};
- \latex{i:\R \to \R}, \latex{i(x)=\left(\frac{1}{3}\right)^{x-4}-3};
- \latex{j:\R \to \R}, \latex{j(x)=\left(\frac{5}{2}\right)^{x}};
- \latex{k:\R \to \R}, \latex{k(x)=2\times 3^{x-3}-6}.
{{exercise_number}}. If at time \latex{0} there had been \latex{N_0} non-decayed atoms in the radioactive material, then in \latex{t} time the number of still non-decayed atoms is \latex{N(t)=N_{0}\times e^{-\lambda \times t}}. \latex{\lambda} is the disintegration constant of the material. The disintegration constant of radium is: \latex{\lambda=4.279 \times 10^{-4}} \latex{\frac{1}{\text {year}}}.
- What ratio of a radium atom decays in \latex{100} years?
- What percent of a radium atom decays in \latex{1620} years?
{{exercise_number}}. Plot the graph of the function \latex{f:\R \backslash \left\{2\right\} \to \R }, \latex{f(x)=3^{\frac{x^2-4}{x-2} }}.
Puzzle
Plot the graph of the function \latex{g: \R \to \R}, \latex{g(x)=5^{\left(x-1\right)^{2}-2x-x^2} }.
