In some cases it is necessary to recover the exponent of a power. Currently, we have no means of expressing the result of this activity. Therefore, we introduce a new operation. This new operation helps us to recover the argument of an exponential function #a^x#. In general, such a recovering operation is called an inverse function.
Let #a\gt0# and #a\ne 1#. Then #\log_a(x) # is the real number #y# defined by the property #a^y#.
The number #\log_a(x)# is called the logarithm of #x# to the base #a#.
The logarithm #\log_a(x)# is indeed the function that assigns to #x# its exponent when written as a power of #a#. Look at the following examples.
- #\log_2(8)=3#, since #2^3=8#
- #\log_{16}(4)=\frac{1}{2}#, since #16^{\frac{1}{2}}=4#
- #\log_2(32)=5#, since #2^5=32#
- #\log_{\frac{1}{2}}(64)=-6#, since #\left(\frac{1}{2}\right)^{-6}=64#
We sometimes use #\log(x)# as an abbreviation of #\log_{10}(x)#.
Others sometimes use it as an abbreviation of #\log_{\ee}(x)#, where #\ee\approx 2.71828#, is Euler's number. To this end, we employ the notation #\ln(x)#, where the symbol 'ln' stands for natural logarithm. In this chapter we will not really use the natural logarithm, nor Euler's number.
If #y\le0#, then there is no value of #x# satisfying #a^x = y#. Therefore #\log_a(x)# is only defined for #x\gt0#.
If #a=1#, then also #a^y=1#. The number #y# with #a^y=x# does not exist if #x\ne1# and is not uniquely defined if #x=1#.
If #f(x)# is a function with the property that, for each #y# there is at most one value #x# satisfying #f(x) =y#, then the function #g(x)# with the property that #y=g(x)# satisfies #f(y) = x# is the inverse function of #f(x)#.
In case of the logarithm we are considering #f(x) =a^x#, the power of #a# with exponent #x#, whose inverse function is #g(x) = \log_a(x)#.
Below we state two important properties of logarithms, which express that #\log_a(x)# is the 'recover' function, or the inverse function, of the exponential function #a^x#.
Let #a\gt0# and #a\ne1#. The logarithm #\log_a(x)# is the inverse function of the exponential function #a^x#. In particular:
- #\log_a(a^x) = x# for all #x#.
- #a^{\log_a(x)} = x# for all #x\gt 0#.
We will deduce the two rules.
1. #\log_a(a^x) = x# for all #x#. According to the definition, #\log_a(x)=y# holds if and only if #x=a^y#. If we substitute #x# by #a^x# and #y# by #x# in this expression then it says: #\log_a(a^x)=x# if and only if #a^x=a^x#. Because the right hand side is always true, the same holds for the left hand side: #\log_a(a^x)=x#.
2. #a^{\log_a(x)} = x# for all #x\gt 0#. According to the definition, #\log_a(x)=y# holds if and only if #x=a^y#. If we substitute #y# by #\log_a(x)# in this expression then it says: #\log_a(x)=\log_a(x)# if and only if #x=a^{\log_a(x)}#. Because the left hand side is always true, the same holds for the right hand side: #x=a^{\log_a(x)}#.
The second statement means that #\log_a(x)# is the inverse function of #a^x#. The first statement means that #a^x# is the inverse function of #\log_a(x)#. This is a well-known property of functions: if #g(x)# is the inverse function of #f(x)#, then #f(x)# is the inverse function of #g(x)#.
Below is the graph of the function #\log_a#. The values of the base #a# can be varied by using the slider.
By moving the slider you will find support for the following statement.
If #a\gt1#, then #\log_a(x)# is increasing, and if #0\lt a\lt1#, it is decreasing.
- #\log_7(7^2) = 2\lt 3 = \log_7(7^3)#
- #\log_{\frac{1}{7}}(7^2) =- 2\gt -3 = \log_{\frac{1}{7}}(7^3)#
Suppose that #x# and #y# are numbers with #0\lt x\lt y#. According to the above theorem #\log_a(x)# and #\log_a(y)# are numbers with #{\log_a(a^x)} = x# en #{\log_a(a^y)} = y#. So #x\lt y# holds if and only if #{\log_a(a^x)}\lt {\log_a(a^y)} #.
Assume first #a\gt 1#. We know about exponentiation that #x\lt y# holds if and only if #a^x\lt a^y#. In view of the above equivalence, we conclude that #{\log_a(a^x)}\lt {\log_a(a^y)} # if and only if #a^x\lt a^y#. Replacing #a^x# by #x# and #a^y# by #y#, we find: #{\log_a(x)}\lt {\log_a(y)} # if and only if #x\lt y#. This means that #\log_a(x)# is increasing.
Assume now that #0\lt a\lt 1#. We know about exponentiation that #x\lt y# if and only if #a^x\gt a^y#. In view of the first equivalence of this proof, we conclude that #{\log_a(a^x)}\lt {\log_a(a^y)}# if and only if #a^x\gt a^y#. Replacing #a^x# by #x# and #a^y# by #y#, we find: #{\log_a(x)}\lt {\log_a(y)} # if and only if #x\gt y#. This means that #\log_a(x)# is decreasing.
Write #\log_{2}\left(16\right)# as an exact number without logarithms.
#\log_{2}\left(16\right)=# #4#
The answer follows from #16=2^{4}# and the rule #\log_a(a^x)=x#:\[\log_{2}\left(16\right)=\log_{2}\left(2^{4}\right)=4\]
The notion of logarithm
