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The concept of the logarithm
The astronomers of the middle ages had to solve numerous calculation exercises till they found out that this work can significantly be simplified when calculating with powers. Later on they also realised that it was enough to write down and calculate with the indices only instead of the powers.
The indices belonging to the distinct bases were given separate names and were tabulated.
Example 1
Bacteria are cultured in a laboratory for the experiments. The number of bacteria doubles every hour. How much time is needed for the number of entities to reach \latex{10,000} which are necessary for the experiment if there are \latex{800} entities at a given moment?
Solution
Since the number if bacteria doubles every hour, after time \latex{x} there will be \latex{800 \times 2^x} bacteria.
We have to solve the equation
\latex{800\times2^x=10,000}.
When dividing by \latex{800}:
\latex{2^x=12.5}
Since the range of the exponential function with base \latex{2} is the set of positive numbers, the equation will have a solution (Figure 17).
Due to the exponential function being strictly increasing it is true that
\latex{3 \lt x \lt 4}.
Let us tabulate the values of \latex{2^x} with the help of a pocket calculator, if \latex{x} is between \latex{3} and \latex{4}:
\latex{x}
\latex{2^x}
\latex{\underbrace{\phantom{000000000000}}}
\latex{3}
\latex{3.1}
\latex{3.2}
\latex{3.3}
\latex{3.4}
\latex{3.5}
\latex{3.6}
\latex{3.7}
\latex{3.8}
\latex{3.9}
\latex{4}
\latex{8}
\latex{8.57}
\latex{9.19}
\latex{9.85}
\latex{10.56}
\latex{11.31}
\latex{12.13}
\latex{12.99}
\latex{13.93}
\latex{14.39}
\latex{16}
\latex{x}
\latex{2^x}
\latex{3.6}
\latex{3.61}
\latex{3.62}
\latex{3.63}
\latex{3.64}
\latex{3.65}
\latex{3.66}
\latex{3.67}
\latex{3.68}
\latex{3.69}
\latex{3.70}
\latex{12.13}
\latex{12.21}
\latex{12.29}
\latex{12.38}
\latex{12.47}
\latex{12.55}
\latex{12.64}
\latex{12.73}
\latex{12.82}
\latex{12.91}
\latex{12.99}
\latex{y}
\latex{x}
\latex{y=2^x}
\latex{12.5}
\latex{1}
\latex{1}
\latex{3}
\latex{4}
Figure 17
The solution of the equation rounded to \latex{2} decimal places
\latex{x \approx 3.64}.
So the number of entities will be appropriate in about \latex{3.64} hours (\latex{\approx 3} hours \latex{38} minutes).
The existing exponent, which in this case can only be determined approximately, is called the logarithm of \latex{12.5} with base two.
DEFINITION: The logarithm of number \latex{b} with base \latex{a} is the exponent what \latex{a} needs to be raised to the power of to get \latex{b}, where \latex{a \gt 0}; \latex{a \ne 1} and \latex{b \gt 0}.
Notation: \latex{\log_a b}.
\latex{a^{\log_ab}=b}
\latex{a \gt 0}; \latex{a \ne 1}; \latex{b \gt 0}
Since for example the equation \latex{2^x = b} only has a solution for positive numbers \latex{b}, the logarithm of \latex{0} and of negative numbers is not defined.
The range of the exponential function with base \latex{1} is the set \latex{\lbrace 1 \rbrace}, no other positive number can be expressed as the power of \latex{1}, therefore the logarithm with base \latex{1} is not defined. (Figure 18)
The notation of the logarithm with base \latex{10} is \latex{\log}.
The notation of the logarithm with base \latex{e} is \latex{\ln}.
\latex{y}
\latex{x}
\latex{y=1^x}
\latex{-3}
Figure 18
\latex{1}
\latex{1}
\latex{y=a^x}
\latex{\left(0 \lt\ a \lt 1\right)}
\latex{y=a^x}
\latex{\left(a \gt 1\right)}
Example 2
Let us calculate the following logarithm values:
- \latex{\log_2 4};
- \latex{\log_3 27};
- \latex{\log_4 4};
- \latex{\ln 1};
- \latex{\log 0.1};
- \latex{\log \left(-10\right)};
- \latex{\log_9 3};
- \latex{\log_4 8}.
Solution
- \latex{2^2 = 4}, therefore \latex{\log_2 4 = \log_2 2^2 = 2};
- \latex{3^3 = 27}, therefore \latex{\log_3 27 = \log_3 3^3 = 3};
- \latex{4^1 = 4}, therefore \latex{\log_4 4 = 1};
- \latex{e^0 = 1}, therefore \latex{\ln 1 = \ln e^0 = 0};
- \latex{10^{-1} = 0.1}, therefore \latex{\log 0.1 = \log 10^{-1} = -1};
- there is no such exponent for which it is true that \latex{10^x = –10}, therefore it does not have a meaning;
- \latex{9^{\frac{1}{2}} = \sqrt{9}=3}, therefore \latex{\log_9 3 = \log_9 9^{\frac{1}{2}} = \frac{1}{2}};
- \latex{8=2^3=\left(4^{\frac{1}{2}}\right)^3=4^{\frac{3}{2}}}, therefore \latex{\log_4 8 = \log_4 4^{\frac{3}{2}}=\frac{3}{2}}.
Example 4
Let us calculate the following powers:
a) \latex{5^{\log_5 7}};
b) \latex{8^{\log_8 15}};
c) \latex{e^{\ln 23}};
d) \latex{3^{\log_3\left( -2\right)}}.
Solution
Based on the definition of logarithm:
- \latex{5^{\log_5 7}=7};
- \latex{8^{\log_8 15}=15};
- \latex{e^{\ln 23}=23}.
- Since logarithms are defined only for positive numbers, the power does not have a meaning. (Figure 19)
\latex{y}
\latex{x}
\latex{-2}
Figure 19
\latex{1}
\latex{1}
\latex{y=3^x}
Example 4
Let us calculate the following powers:
- \latex{7^{2 \times \log_7 3}};
- \latex{100^{\log 4}};
- \latex{2^{\log_4 9}};
- \latex{\left(\frac{1}{3}\right)^{\log_3 5}}.
Solution
- Because of the product in the exponent let us rearrange it:
\latex{7^{2 \times \log_7 3}=\left( 7^{\log_7 3}\right)^2 = 3^2 = 9}.
- Let us express the base of the power with the help of the base of the logarithm:
\latex{100^{\log 4}=\left( 10^2\right)^{\log 4}=10^{2 \times \log 4}=\left( 10^{\log 4}\right)^2=4^2=16}.
- Let us express the base of the power as a power of \latex{ 4 }:
\latex{2^{\log_4 9}=\left( 4^{\frac{1}{2}}\right)^{\log_4 9}=4^{\frac{1}{2}\times \log_4 9}=\left( 4^{\log_4 9}\right)^{\frac{1}{2}}=9^{\frac{1}{2}}=3}.
- By applying the previous method:
\latex{\left( \frac{1}{3} \right)^{\log_3 5}=\left( 3^{-1}\right)^{\log_3 5}=3^{\left(-1\right) \times \log_3 5}=\left( 3^{\log_35}\right)^{-1}=5^{-1}=\frac{1}{5}}.
Example 5
Let us determine the values of the unknowns:
- \latex{\log_3 a = 2};
- \latex{\log_5 b = 3};
- \latex{\log_4 c = -2};
- \latex{\log_9 x = 0};
- \latex{\ln d = 0.5};
- \latex{\log y = -3};
- \latex{\log_0 f = 4};
- \latex{\log_{100} h = -\frac{1}{2}}.
Solution
According to the definition of the logarithm:
- \latex{a = 3^2 = 9};
- \latex{b = 533 = 125};
- \latex{c = 4^{-2}=\frac{1}{16}};
- \latex{x = 9^0 = 1};
- \latex{d = e^{0.5} = \sqrt{e}\approx 1.649};
- \latex{y = 10^{-3}=\frac{1}{1,000}=0.001};
- The base can only be a positive number; the expression does not have a meaning;
- \latex{h= 100^{-\frac{1}{2}}=\sqrt{100^{-1}}=\sqrt{\frac{1}{100}}=\frac{1}{10}}.
Example 6
Let us calculate the bases of the logarithms:
- \latex{\log_a 27=3};
- \latex{\log_b 7=0.5};
- \latex{\log_c 5=-1}.
Solution
Based on the definition of logarithm:
a) \latex{a^3=27}, \latex{a=3; \qquad\quad} b) \latex{b^{0.5}=7}, \latex{b=49;\qquad\quad} c) \latex{c^{-1}=5}, \latex{c=\frac{1}{5}}.
Example 7
Let us calculate the bases of the logarithms:
- \latex{\log_a\left( -2 \right)=1};
- \latex{\log_b 12=5};
- \latex{log_c \frac{49}{25}=-2};
- \latex{\log_d 1=0}.
Solution
- the logarithm was defined only for positive numbers, therefore the equation does not have a solution;
- \latex{b^5=12}, \latex{b=\sqrt[5]{12}};
- \latex{c^{-2}=\frac{49}{25}}, \latex{c=\frac{5}{7}};
- \latex{d^0=1}, \latex{d \in \R}; \latex{d \gt0}; \latex{d \ne 1}.
Example 8
Let us give the largest subset of the set of real numbers on which the expressions below have a meaning:
- \latex{\log_5\left(x-2 \right)};
- \latex{\log_3 \left(7x+23\right)};
- \latex{log_7 \left( x+2 \right)-\log_7 x.}
Solution
The expressions have a meaning if and only if the corresponding expressions are positive.
- \latex{x-2\gt0}, thus \latex{x\gt 2};
- \latex{7x + 23 \gt 0}, thus \latex{x \gt -\frac{23}{7}};
- \latex{x+2\gt0}, \latex{x\gt-2} and \latex{x\gt0}. The common solution is: \latex{x\gt0}.
Example 9
Let us give the largest subset of the set of real numbers on which the expression \latex{\log\frac{x-4}{3-x}} has a meaning.
Solution
\latex{\frac{x-4}{3-x}\gt0} is satisfied if the numerator and the denominator have the same sign.
Case II: Both the numerator and the
denominator are negative:
denominator are negative:
\latex{x-4\gt0}, \latex{x\gt4},
\latex{3-x\gt0}, \latex{3\gt x}.
\latex{3-x\gt0}, \latex{3\gt x}.
\latex{3}
\latex{4}
\latex{3}
\latex{4}
Case I: Both the numerator and the
denominator are positive:
denominator are positive:
We obtain no solution.
The solution: \latex{3 \lt x \lt 4}.
\latex{x-4\lt0}, \latex{x\lt4},
\latex{3-x\lt0}, \latex{3\lt x}.
\latex{3-x\lt0}, \latex{3\lt x}.
So the expression can be defined in the case when \latex{3 \lt x \lt 4}.
Example 10
Let us give the largest subset of the set of real numbers on which the expressions below have a meaning:
- \latex{\log_x\left(x^2-11x+24\right)};
- \latex{\log_2 \left(3^{x-1}-243\right)};
- \latex{log_{11} \frac{23-3^x}{x-1}};
- \latex{log \left(-t^4+5t^2-4\right)}.
Based on the definition we know that the independent variable of the logarithm has to be positive, and its base has to be a positive value different from \latex{1}.
Solution (a)
Thus
\latex{x^2-11x+24 \gt0\quad} and \latex{\quad x \gt 0}; \latex{x \ne 1};
The zeros of the quadratic expression are: \latex{x_1 = 3}; \latex{x_2 = 8}. The solution of the inequality is: \latex{x \lt 3} or \latex{x \gt 8}. By comparing to the other condition the solution is (Figure 20):
\latex{0 \lt x \lt3}; \latex{x \ne 1\quad} or \latex{\quad x \gt 8}.
Solution (b)
The expression can be defined if
\latex{3^{x-1}-243 \gt 0}
\latex{\quad\qquad3^{x-1} \gt 3^5};
\latex{y}
\latex{x}
\latex{y=x^2-11x+24}
\latex{1}
\latex{3}
\latex{1}
\latex{8}
Figure 20
Solution (c)
\latex{\frac{23-3^x}{x-1}\gt0} is satisfied if the numerator and the denominator have the same sign.
Case II: Both the numerator and the
denominator are negative:
denominator are negative:
\latex{23-3^x\gt0 \Harr 23\gt 3^x},
thus: \latex{\log_3 23\gt x};
thus: \latex{\log_3 23\gt x};
Case I: Both the numerator and the
denominator are positive:
denominator are positive:
\latex{23-3^x\lt0 \Harr 23 \lt 3^x},
thus: \latex{\log_3 23 \lt x};
thus: \latex{\log_3 23 \lt x};
\latex{x-1\gt 0}, which implies \latex{x\gt1}.
Since \latex{\log_3 23 \gt 1}, the solution is:
\latex{1 \lt x \lt \log_3 23}.
\latex{1 \lt x \lt \log_3 23}.
\latex{x-1\lt 0}, which implies \latex{x\lt1}.
In this case there is no solution.
The solution from case I is: \latex{1 \lt x \lt log_3 23}.
Solution (d)
The expression can be defined if \latex{–t^4+5t^2-4\gt0}. Let us introduce a new variable to solve the inequality of degree four which can be reduced to a quadratic inequality: let \latex{x = t^2}.
The inequality is:
\latex{-x^2 + 5x -4 \gt 0},
the zeros of the quadratic expression are: \latex{x_1 = 1}, \latex{x_2 = 4}; the solution of the quadratic inequality (Figure 21):
\latex{1 \lt x \lt 4}.
By substitution:
\latex{1 \lt t^2 \lt 4},
\latex{1 \lt |t| \lt 2},
\latex{1 \lt |t| \lt 2},
and the solution of this is:
\latex{y}
\latex{x}
\latex{y=-x^2+5x-4}
\latex{1}
\latex{4}
\latex{1}
Figure 21
\latex{-2 \lt t \ -1} or \latex{1 \lt t \lt 2}.
Exercises
{{exercise_number}}. Calculate the values of the logarithms:
- \latex{\log_3 9};
- \latex{\log_9 3};
- \latex{\log_2 1,024};
- \latex{\log_5 625};
- \latex{\log 100,000};
- \latex{\log_{\frac{3}{4}}\frac{16}{9}};
- \latex{\log_{\frac{1}{3}}27};
- \latex{\log_8 8}.
{{exercise_number}}. Calculate the following powers:
- \latex{4^{\log_4 11}};
- \latex{10^{\log 3}};
- \latex{e^{\ln 5}};
- \latex{11^{\log_{\frac{1}{11}}3}};
- \latex{7^{\log_{49}16}};
- \latex{3^{\log_9 25}};
- \latex{9^{\log_3 25}};
- \latex{\left(\frac{1}{2}\right)^{\log_2 3}}.
{{exercise_number}}. Calculate the unknowns:
- \latex{\log_5 a=1};
- \latex{\log_7 b=2};
- \latex{\log c =0};
- \latex{\ln d=-3};
- \latex{\log_{\frac{1}{5}} e=2};
- \latex{\log_8 f=-\frac{2}{3}};
- \latex{\log_{\frac{2}{3}}g = -2};
- \latex{\log_{\frac{3}{7}}h = -\frac{1}{2}}.
{{exercise_number}}. Calculate the bases of the logarithms:
- \latex{\log_a 9=2};
- \latex{\log_b 10,000=4};
- \latex{\log_c 5 =0.5};
- \latex{\log_d 3=-\frac{1}{2}};
- \latex{\log_e 4=-1};
- \latex{\log_f 25=-\frac{3}{2}};
- \latex{\log_g \frac{9}{4}= -2};
- \latex{\log_h\left(-7\right) = 2}.
{{exercise_number}}. Give the largest subset of the set of real numbers on which the following expressions have a meaning:
- \latex{\log_3 \left(2x-3 \right)+\log_3 \left(x-1\right)};
- \latex{\log_4 \left( 5x-1\right)+\log_4 \left( 2-x\right)};
- \latex{\log \left(x^2-7x+12\right)-\log \left(2x-7\right)};
- \latex{\log_x\left(5-2x\right)};
- \latex{\log_8\frac{3^x-9}{5x-2}};
- \latex{\log_2\left(5-|x|\right)\times \log_2\left(x^2-16\right)}.
{{exercise_number}}. If at time \latex{0} there had been \latex{N_0} non-decayed atoms in the radioactive material, then at time \latex{t} the number of still non-decayed atoms will be \latex{N\left(t\right) = N_0 \times e^{-\lambda\times t}\times \lambda} is the disintegration constant of the material. The disintegration constant of radium is: \latex{\lambda = 4.279 \times 10^{-4}\frac{1}{\text{year}}}. How much time is needed for the radium atom to have decayed by half? (This time is called the decay half-life of radioactive materials.)