Logarithm laws enable several logarithmic calculations to be performed without a calculator. Logarithm laws are frequently used in financial mathematics for solving situations involving compound interest, and in physics to calculate a half-life.
|
Special Cases |
|
|---|---|
|
|\log_c 1 =0| |
|
|
|\log_c c =1| |
|
|
Logarithm Where the Argument Is Equal to the Base With an Exponent |
|\log_c c^t = t| |
|
Laws |
|
|
|\log_c(M \times N) = \log_c M + \log_c N| |
|
|
|\log_{c}\dfrac{M}{N}=\log_{c}M-\log_{c}N| |
|
|
|\log_c M^{\large n} = n \log_c M| |
|
|
|\log_{\large\frac{_{1}}{c}}M=-\log_{c}M| |
|
|
|\log_{c}M=\dfrac{\log_{a}M}{\log_{a}c}| |
|
Note:
-
For each law, we have |\{c,a,M,N \} \in\ ]0,+\infty[| and |n \in \mathbb{R}.|
-
The laws can be read both from left to right and from right to left. The laws can be used both ways, depending on the problem.
A logarithm where the argument is |1| is always |0.| ||\log_c 1=0||
What is the value of |\log_8 1?|
Determining |\log_8 1| amounts to asking: “What exponent should be applied to |8| to obtain |1?|”
||\begin{align}\log_{\color{#3b87cd}{8}}\color{#ec0000}{1}=\ \color{#3a9a38}{?}\quad\Leftrightarrow\quad\color{#3b87cd}{8}^\color{#3a9a38}{?}&=\color{#ec0000}{1}\\ \color{#3b87cd}{8}^\color{#3a9a38}{0}&=\color{#ec0000}{1}\end{align}||
Answer: Since |0| is the exponent that must be applied to |8| to obtain |1,| we conclude that |\log_8 1=0.|
||\color{#3b87cd}{c}^\color{#3a9a38}{0}=\color{#ec0000}{1}\quad\Leftrightarrow\quad\log_ {\color{#3b87cd}{c}}\color{#ec0000}{1}=\color{#3a9a38}{0}||
Any base raised to the exponent |0| is |1| unless the base is |0.|
A logarithm where the base is the same as the argument is equal to |1.| ||\log_c c=1||
What is the value of |\log_{12} 12?|
Determining |\log_{12} 12| amounts to asking: “What exponent should be applied to |12| to obtain |12?|”
||\begin{align}\log_{\color{#3b87cd}{12}}\color{#ec0000}{12}=\ \color{#3a9a38}{?}\quad\Leftrightarrow\quad\color{#3b87cd}{12}^\color{#3a9a38}{?}&=\color{#ec0000}{12}\\ \color{#3b87cd}{12}^\color{#3a9a38}{1}&=\color{#ec0000}{12}\end{align}||
Answer: Since |1| is the exponent that must be applied to |12| to obtain |12,| we conclude that |\log_{12} 12=1.|
||\color{#3b87cd}{c}^\color{#3a9a38}{1}=\color{#ec0000}{c}\quad\Leftrightarrow\quad\log_{ \color{#3b87cd}{c}}\color{#ec0000}{c}=\color{#3a9a38}{1}||
Any base given an exponent of |1| is the base itself.
This case is the direct result of the change from logarithmic notation to exponential notation.
||\log_c c^t=t||
What is the value of |\log_5 125?|
We know that we can express the number |125| as a power of |5.|
||125=5^3||
Answer:
||\begin{align}\log_\color{#3b87cd}5 \color{#ec0000}{125}&=\log_\color{#3b87cd}5 \color{#ec0000}{5^3}\\&=\color{#3a9a38}3\end{align}||
|\color{#ec0000}t| is the exponent that must be assigned to |c| to obtain |c^t.| ||c^\color{#ec0000}t=c^t \leftrightarrow\ \log_c c^t=\color{#ec0000}t||
When the argument of the logarithm consists of 2 factors multiplied together, it can be written as the addition of 2 logarithmic expressions.
||\log_c (M\times N)=\log_c M+\log_c N||
Note: The value of the base does not change when this law is applied.
Example 1
What is the value of |\log_{12} 4+\log_{12} 36?|
The following equality is obtained by applying the product law for logarithms. ||\begin{align}\log_{12}4+\log_{12}36&=\log_{12} (4 \times 36) \\&= \log_{12} (144) \end{align}||
Next, we examine the problem: “What exponent should be applied to |12| to obtain |144?|”
||\begin{align}\log_{\color{#3b87cd}{12}}\color{#ec0000}{144}=\ \color{#3a9a38}{?}\quad\Leftrightarrow\quad\color{#3b87cd}{12}^\color{#3a9a38}{?}&=\color{#ec0000}{144}\\ \color{#3b87cd}{12}^\color{#3a9a38}{2}&=\color{#ec0000}{144}\end{align}||
The exponent that must be applied to |12| to obtain |144| is |2.|
Answer:
||\begin{align}\log_{12}4+\log_{12}36&= \log_{12} (144)\\&= 2\end{align}||
Example 2
Decompose the following expression into a sum of logarithms: |\log_{10} 15.|
We know that |15=3\times5.|
Use the product law for logarithms to decompose the expression.
Answer: ||\begin{align}\log_{10}15&=\log_{10}(3\times5)\\&=\log_{10}3+\log_{10}5\end{align}||
Note: Decomposition is useful for simplifying expressions when performing logarithmic calculations.
Let |x=\log_c M| and |y=\log_c N.|
|x| is the exponent that must be applied to |c| to obtain |M,| and |y| is the exponent that must be applied to |c| to obtain |N.| ||\begin{align}x&=\log_cM&\Leftrightarrow&&c^x&=M\\y&=\log_c N&\Leftrightarrow&&c^y&=N\end{align}||
When the terms are multiplied together, the following result is obtained.
||c^x \times c^y=M \times N||
Apply the product law for exponentials. ||c^{x+y}=M \times N||
With this equality, switch to logarithmic notation. The exponent to be applied to |c| to obtain |M \times N| is |x+y.|
||c^{x+y}=M \times N\ \Leftrightarrow \ \log_c {(M \times N)}=x+y||
It was initially assumed that |x=\log_c M| and |y=\log_c N,| so we replace them in the previous equation. The result is the product law for logarithms.
||\begin{align}\log_c {(M \times N)}&=\color{#fa7921}x+\color{#ff55c3}y\\ \log_c {(M \times N)}&=\color{#fa7921}{\log_c M}+\color{#ff55c3}{\log_c N}\end{align}||
When the argument of the logarithm consists of a term divided by another, it can be written as the subtraction of 2 logarithmic expressions. ||\log_c \left(\dfrac{M}{N}\right)=\log_c M-\log_c N||
Note: The value of the base does not change when this law is used. Also, the order of the arguments must be respected.
Example 1
What is the value of |\log_2 320-\log_2 5?|
Apply the quotient law for logarithms to obtain the following equality.
||\begin{align} \log_2 320-\log_2 5&=\log_2 \left(\dfrac{320}{5}\right)\\&=\log_2 64\end{align}||
Next, examine the problem: “What exponent should be applied to |2| to obtain |64?|”
||\begin{align}\log_{\color{#3b87cd}{2}}\color{#ec0000}{64}=\ \color{#3a9a38}{?}\quad\Leftrightarrow\quad\color{#3b87cd}{2}^\color{#3a9a38}{?}&=\color{#ec0000}{64}\\ \color{#3b87cd}{2}^\color{#3a9a38}{6}&=\color{#ec0000}{64}\end{align}||
The exponent that must be applied to |2| to obtain |64| is |6.|
Answer:
||\begin{align}\log_2 320-\log_2 5&=\log_2 64\\&=6\end{align}||
Example 2
What is the value of |\log_4 \left(\dfrac{1}{16}\right)?|
Applying the quotient law for logarithms results in the following equality: ||\log_4\left({\dfrac{1}{16}}\right)=\log_4 1-\log_4 16||
Therefore, we want to know which exponent must be applied to |4| to obtain |1| and |16.|
||\begin{align}\log_{\color{#3b87cd}{4}}\color{#ec0000}{1}=\ \color{#3a9a38}{?}\quad\Leftrightarrow\quad\color{#3b87cd}{4}^\color{#3a9a38}{?}&=\color{#ec0000}{1}\\ \color{#3b87cd}{4}^\color{#3a9a38}{0}&=\color{#ec0000}{1}\end{align}||
||\begin{align}\log_{\color{#3b87cd}{4}}\color{#ec0000}{16}=\ \color{#3a9a38}{?}\quad\Leftrightarrow\quad\color{#3b87cd}{4}^\color{#3a9a38}{?}&=\color{#ec0000}{16}\\ \color{#3b87cd}{4}^\color{#3a9a38}{2}&=\color{#ec0000}{16}\end{align}||
The exponents that must be given to |4| to obtain |1| and |16| are |0| and |2,| respectively.
Answer:
||\begin{align}\log_4\left(\dfrac{1}{16}\right)&=\log_4 1-\log_4 16\\&=0-2\\&=-2\end{align}||
Several methods can be used to solve the same problem. For example, the previous problem can be solved by using the logarithm whose argument is equal to the base with an exponent.
||\begin{align}\log_4\left(\dfrac{1}{16}\right)&=\log_4\left(\dfrac{1}{4^2}\right)\\ &=\log_4\left(4^{-2}\right)\\&=-2\end{align}||
Let |x=\log_c M| and |y=\log_c N.|
|x| is the exponent that must be applied to |c| to obtain |M| and |y| is the exponent that must be applied to |c| to obtain |N.|
||\begin{align}x=\log_c M\ &\Leftrightarrow \ c^x=M\\y=\log_c N\ &\Leftrightarrow \ c^y=N\end{align}||
If we divide the 2 terms, we obtain the following equation.
||\dfrac{c^x}{c^y}=\dfrac{M}{N}||
Apply the quotient law for exponents.
||c^{x-y}=\dfrac{M}{N}||
From this equality, switch to logarithmic notation. The exponent that must be applied to |c| to obtain |\dfrac{M}{N}| is |x-y.|
||c^{x-y}=\dfrac{M}{N}\ \Leftrightarrow \ \log_c {\left(\dfrac{M}{N}\right)}=x-y||
It was initially assumed that |x=\log_c M| and |y=\log_c N,| so we replace them in the expression. The result is the quotient law for logarithms.
||\begin{align}\log_c {\left(\dfrac{M}{N }\right)}&=\color{#fa7921}x-\color{#ff55c3}y\\ \log_c {\left(\dfrac{M}{N }\right)}&=\color{#fa7921}{\log_c M}-\color{#ff55c3}{\log_c N}\end{align}||
When the argument of a logarithm is raised to a power, the exponent can be transformed into the coefficient of the same logarithm.
||\log_c {M^n}=n\log_c M||
Note: The value of the base does not change when the law is used.
Example 1
What is the value of |\log_7 49^2?|
Use the power law for logarithms to rewrite this expression.
||\log_\color{#3b87cd}{7} \color{#ec0000}{49^2}=\color{#ec0000}2\log_ \color{#3b87cd}{7} \color{#ec0000}{49}||
Therefore, we are trying to determine which exponent should be applied to |7| to obtain |49.|
||\begin{align}\log_{\color{#3b87cd}{7}}\color{#ec0000}{49}=\ \color{#3a9a38}{?}\quad\Leftrightarrow \quad\color{#3b87cd}{7}^\color{#3a9a38}{?}&=\color{#ec0000}{49}\\ \color{#3b87cd}{7}^\color{#3a9a38} {2}&=\color{#ec0000}{49}\end{align}||
The exponent that must be applied to |7| to obtain |49| is |2.|
Answer: ||\begin{align}\log_7 49^2&=2\log_7 49\\&=2\times2\\&=4\end{align}||
Example 2
What is the value of |2\log_4 8?|
Use the power law for logarithms to rewrite the expression:
||\begin{align} 2\log_\color{#3b87cd}{4} 8&=\log_\color{#3b87cd}{4} \color {#ec0000}{8^2}\\&=\log_\color{#3b87cd}{4} \color{#ec0000}{64}\end{align}||
Therefore, we are trying to determine which exponent should be applied to |4| to obtain |64.|||\begin{align}\log_{\color{#3b87cd}{4}}\color{#ec0000}{64}=\ \color{#3a9a38}{?}\quad\Leftrightarrow \quad\color{#3b87cd}{4}^\color{#3a9a38}{?}&=\color{#ec0000}{64}\\ \color{#3b87cd}{4}^\color{#3a9a38} {3}&=\color{#ec0000}{64}\end{align}||
The exponent that must be applied to |4| to obtain |64| is |3.|
Answer: ||\begin{align}2\log_4 8 &=\log_4 64\\&=3\end{align}||
Let |x=\log_c M.|
|x| is the exponent that must be applied to |c| to obtain |M.|||x=\log_c M\ \Leftrightarrow \ c^x=M||
If both sides of the equality are applied an exponent of |n|, the following is obtained.||(c^x)^{^{\Large{n}}}=M^n||
Apply the power of a power law for exponents. ||c^{xn}=M^n||
From this equality, switch to logarithmic notation. The exponent that must be applied to |c| to obtain |M^n| is |xn.| ||c^{xn}=M^n\ \Leftrightarrow \ \log_c M^n=xn||
It was initially assumed that |x=\log_c M,| so we replace it in the equality. The result is the power law for logarithms.
||\begin{align}\log_c M^n&=\color{#fa7921}xn\\ \log_c M^n&=\color{#fa7921}{\log_c M} \times n\\ \log_c M^n&=n\log_c M\end{align}||
A logarithm where the base is a fraction |\dfrac{1}{c}| is equivalent to the negative of the logarithm with the same argument where the base is |c.|
||\log_{\large\frac{1}{c}}M=-\log_c M||
What is the value of |\log_{\large\frac{1}{3}} 81?|
Apply the fractional base law for logarithms to obtain the equality below.
||\log_\color{#3b87cd}{\large\frac{1}{3}} \color{#ec0000}{81}=-\log_ \color{#3b87cd}{3} \color{#ec0000}{81}||
Next, determine what exponent should be applied to |3| to obtain |81.|
||\begin{align}\log_{\color{#3b87cd}{3}}\color{#ec0000}{81}=\ \color{#3a9a38}{?}\quad\Leftrightarrow\quad\color{#3b87cd}{3}^\color{#3a9a38}{?}&=\color{#ec0000}{81}\\ \color{#3b87cd}{3}^\color{#3a9a38}{4}&=\color{#ec0000}{81}\end{align}||
The exponent that must be applied to |3| to obtain |81| is |4.|
Answer: ||\begin{align}\log_{\large\frac{1}{3}} 81&=-\log_3 81\\&= -4\end{align}||
Let |x=\log_{\large\frac{1}{c}} M.|
|x| is the exponent that must be applied to |\dfrac{1}{c}| to obtain |M.|
||x=\log_{\large\frac{1}{c}} M\ \Leftrightarrow \ \left(\dfrac{1}{c}\right)^x=M||
Apply the negative exponential law for exponents. ||c^{-x}=M||
From this equality, switch to logarithmic notation. The exponent that must be applied to |c| to obtain |M| is |-x.|||c^{-x}=M\ \Leftrightarrow \ -x=\log_c M||
It was initially assumed that |x=\log_{\large\frac{1}{c} } M,| so we replace it in the equation. The result is the fractional base law for logarithms.
||\begin{align}-\color{#fa7921}x&=\log_c M\\-\color{#fa7921}{\log_{\large\frac{1}{c}} M}&=\log_c M\\ \log_{\large\frac{1}{c}} M&=-\log_c M\end{align}||
The logarithm of an argument is equivalent to the logarithm of the same argument divided by the logarithm of its base, provided that the bases of the denominator and numerator are identical.
||\log_c M=\dfrac{\log_a M}{\log_a c}||
where |a\not=0| and |a\not=1|
Note: The order in which the elements are presented in the quotient must be respected. The logarithm of the argument is placed in the numerator and the logarithm of the base is placed in the denominator.
What is the value of |\log_{16} 128?|
Note that |16| and |128| are powers of |2.| Therefore, use the change of base law for logarithms to obtain the following equality.
||\log_\color{#3b87cd}{16} \color{#ec0000}{128}=\dfrac{\log_2 \color{#ec0000}{128}}{\log_2 \color{#3b87cd}{16}}||
Determine the following: “What exponents should be applied to |2| to obtain |128| and |16?|”
||\begin{align}\log_{\color{#3b87cd}{2}}\color{#ec0000}{128}=\ \color{#3a9a38}{?}\quad\Leftrightarrow\quad\color{#3b87cd}{2}^\color{#3a9a38}{?}&=\color{#ec0000}{128}\\ \color{#3b87cd}{2}^\color{#3a9a38}{7}&=\color{#ec0000}{128}\end{align}||
||\begin{align}\log_{\color{#3b87cd}{2}}\color{#ec0000}{16}=\ \color{#3a9a38}{?}\quad\Leftrightarrow\quad\color{#3b87cd}{2}^\color{#3a9a38}{?}&=\color{#ec0000}{16}\\ \color{#3b87cd}{2}^\color{#3a9a38}{4}&=\color{#ec0000}{16}\end{align}||
The exponents that must be applied to |2| to obtain |128| and |16| are |7| and |4,| respectively.
Answer:
||\begin{align} \log_{16} 128&=\dfrac{\log_2 128}{\log_2 16}\\&=\dfrac{7}{4}\end{align}||
Despite the fact that we chose to use base |2| in the previous example, usually base |10| or base |e| are chosen. The majority of calculators are programmed to calculate logarithms in base |10| (log key) or natural logarithms (ln key).
Use a calculator to determine the approximate value of the expression |\log_3 5.|
The expression must be transformed to obtain a logarithm with base |10.| Use the change of base law for logarithms to obtain the following.
||\log_\color{#3b87cd}3 \color{#ec0000}5 = \dfrac{\log_{10}\color{#ec0000}5}{\log_{10} \color{#3b87cd}3}||
The expression can be found using a calculator. Therefore, we have:
||\begin{align}\log_\color{#3b87cd}3 \color{#ec0000}5&=\dfrac{\log_{10}\color{#ec0000}5}{\log_{10} \color{#3b87cd}3}\\&\approx \color{#3a9a38}{1.46}\end{align}||
Note: The natural logarithm can also be used.
||\begin{align}\log_\color{#3b87cd}3 \color{#ec0000}5&= \dfrac{\ln\color{#ec0000}5}{\ln\color{#3b87cd}3}\\&\approx\color{#3a9a38}{1.46}\end{align}||
Let |x=\log_c M.|
|x| is the exponent that must be applied to |c| to obtain |M.| ||x=\log_c M\ \Leftrightarrow \ c^x=M||
The following is obtained.
||\log_a c^x=\log_a M||
Indeed, if 2 expressions are equivalent, then the exponents that must be applied to a certain base |a| to obtain the 2 expressions are also equivalent. From the equality, use the power law for logarithms. ||\begin{align} \log_a c^x &=\log_a M \\ x\log_a c &= \log_a M \end{align}||
Isolate |x.| ||x=\dfrac{\log_a M}{\log_a c}||
It was initially assumed that |x=\log_c M,| so we replace it in the expression. The result is the change of base law for logarithms. ||\begin{align}\color{#fa7921}x&=\dfrac{\log_a M}{\log_a c}\\\color{#fa7921 }{\log_c M}&=\dfrac{\log_a M}{\log_a c}\end{align}||
Pour valider ta compréhension des lois des logarithmes et des exposants de façon interactive, consulte la MiniRécup suivante :
