21.12 - Logarithmic Functions

In the previous section, we completed a discussion on exponential functions. In this section, we will see logarithmic functions.

Some basic details about logarithmic functions can be written in 3 steps:
1. Consider the exponential function that we saw in the previous section: y = f(x) = b^x.
♦ where b > 0 and b ≠ 1.
2. Here "b" is a constant. x and y are variables.
• If we give an input x, we will get an output y.
• For example, if b = 2 and x = 3, we get: y = 2³ = 8.
• In such a situation, we say that:
Logarithm of 8 to the base 2 is 3
• In general, if y = b^x, we say that:
Logarithm of y to the base b is x.
3. Logarithm of y to the base b is denoted as log_b y.
• So if y = b^x, we can write: log_b y = x.

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Let us see some solved examples:

Solved example 21.41
Evaluate the following logarithms:
(a) log₁₀ 1000 (b) log₄ 16 (c) log₅ 625 (d) log_1/6 36 (e) $\log_9 \frac{1}{531441}$ (f) $\log_{\frac{3}{2}} \frac{27}{8}$
Solution:
Part (a)
1. Let log₁₀ 1000 = x
2. Then we can write: 1000 = 10^x.
3. We have: 1000 = 10³.
4. From (2) and (3), we get: 10^x = 10³.
• In the above equation, bases are the same. So equating the powers, we get: x = 3

Part (b)
1. Let log₄ 16 = x
2. Then we can write: 16 = 4^x.
3. We have: 16 = 4².
4. From (2) and (3), we get: 4^x = 4².
• In the above equation, bases are the same. So equating the powers, we get: x = 2

Part (c)
1. Let log₅ 625 = x
2. Then we can write: 625 = 5^x.
3. We have: 625 = 5⁴.
4. From (2) and (3), we get: 5^x = 5⁴.
• In the above equation, bases are the same. So equating the powers, we get: x = 4

Part (d)
1. Let $\log_{\frac{1}{6}} 36$ = x
2. Then we can write: $36 = \left(\frac{1}{6} \right)^x$.
⇒ $36 = 6^{-x}$
3. We have: 36 = 6².
4. From (2) and (3), we get: 6^−x = 6².
• In the above equation, bases are the same. So equating the powers, we get: x = −2

Part (e)
1. Let $\log_9 \frac{1}{531441}$ = x
2. Then we can write: $\frac{1}{531441}$ = 9^x.
⇒ (531441)⁻¹ = 9^x.
3. We have: 531441 = 9⁶.
⇒ (531441)⁻¹ = (9⁶)⁻¹ = 9⁻⁶.
4. From (2) and (3), we get: 9^x = 9⁻⁶.
• In the above equation, bases are the same. So equating the powers, we get: x = −6

Part (f)
1. Let $\log_{\frac{3}{2}} \frac{27}{8}$ = x
2. Then we can write: $\frac{27}{8} = \left(\frac{3}{2} \right)^x$.
3. We have: $\frac{27}{8} = \left(\frac{3}{2} \right)^3$.
4. From (2) and (3), we get: $\left(\frac{3}{2} \right)^x = \left(\frac{3}{2} \right)^3 $.
• In the above equation, bases are the same. So equating the powers, we get: x = 3

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Now we will see how logarithm can be used as a function. It can be written in 4 steps:
1. We know that, if y = b^x, then: log_b y = x.
2. Consider the expression log_b y = x.
• Here, the input y is being processed to obtain y.
• In other words, y is being subjected to a process. The process is nothing but "finding the logarithm of y". (The solved examples that we saw just above, show us how to find the logarithm of a given number). The resulting logarithm is the output.
3. But for functions,
    ♦ Input values are denoted as x.
      ✰ They are plotted along the x-axis.
    ♦ Output values are denoted as y.
      ✰ They are plotted along the y-axis
• So in the expression log_b y = x, we need to interchange x and y. We get: log_b x = y.
4. So we get a function in which, input x is processed to give output y.
• The process is nothing but "finding the logarithm of x". The resulting logarithm is the output y.
• We can write: y = f(x) = log_b x.
This is called logarithmic function.

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Let us write the important features about logarithmic functions. It can be written in 7 steps:
1. The general form of the logarithmic function is: f(x) = log_b x.
   ♦ b should be greater than zero.
   ♦ b should not be equal to 1.

2. Let us see why b should be greater than zero. It can be written in (iii) steps.
(i) Suppose that, b = −2.
Then the function will be y = f(x) = log₍₋₂₎ x
(ii) While plotting the graph, when the input is 2, we need to find log₍₋₂₎ 2
• Let us try:
   ♦ Assume log₍₋₂₎ 2 = y
   ♦ Then we can write: 2 = (−2)^y.
   ♦ ⇒ −2 = 2^(1/y).
   ♦ This is impossible because, no power of 2 will give a negative value.
(iii) To avoid such situations, we avoid −ve numbers altogether.

3. Let us see why b should not be equal to one. It can be written in (iii) steps.
(i) Suppose that, b = 1.
Then the function will be y = f(x) = log₁ x
(ii) While plotting the graph, when the input is 2, we need to find log₁ 2
• Let us try:
   ♦ Assume log₁ 2 = y
   ♦ Then we can write: 2 = (1)^y.
   ♦ This is impossible because, all powers of 1 give 1.
(iii) To avoid such a situation, we avoid 1.

4. Fig.21.18 below shows the graphs of some simple logarithmic functions.

Fig.21.18

• Red, and yellow belong to the category: b > 1
• Green and magenta belong to the category: 0 < b < 1
• White shows b = e. It belong to the category: b > 1

5. From the graphs, we see that,
• When b >1:
   ♦ As x increases towards ∞, f(x) also approaches ∞.
         ✰ Red, yellow and white are rising up.
   ♦ As x decreases towards zero, f(x) approaches −∞.
         ✰ Red, yellow and white are falling down.
• When 0 < b < 1:
    ♦ As x increases towards ∞, f(x) approaches −∞.
          ✰ Green and magenta are falling down.
    ♦ As x decreases towards zero, f(x) approaches ∞.
          ✰ Green and magenta are rising up.


6. From the graphs, we see that:
• Input x cannot be a −ve number.
The reason can be demonstrated in (iii) steps:
(i) Suppose that, x = −1000 and b = 10
(Recall that, b cannot be −ve)
So we want to find y, where y = log₁₀ (−1000)
(ii) Let us try:
   ♦ We can write: −1000 = (10)^y.
   ♦ This is impossible because, no power of 10 will give a negative value.
(iii) So for logarithmic functions, input can never be −ve.
We will see the actual proof in higher classes.

• Since no input can be −ve, we say that:
Domain of the logarithmic function is R⁺.

7. From the graphs, we get the following information also:
• The output of a log function can never be zero.
   ♦ Note that, none of the graphs touch the y-axis.
• The range of a log function is (−∞,∞).
• For a log function, the point (1,0) will be always available.
   ♦ This is because, any number raised to the power zero, is 1. Which implies: Whatever be the base, logarithm of 1 is zero.

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Now we will discuss the fact that, log function is the inverse of the exponential function. It can be written in 4 steps:
1. Fig.21.19 shows the graphs of three functions:
(i) y = f(x) = e^x. (red color)
(ii) y = f(x) = x. (magenta color)
(iii) y = f(x) = log_e x. (yellow color)

Fig.21.19

2. Mark any point P on the magenta line.
• Through P, draw a white dashed line perpendicular to the magenta line.
• The white dashed line,
   ♦ intersects the red curve at Q.
   ♦ intersects the yellow curve at R.
• The distances PQ and PR are equal.

3. PQ and PR are equal because,
   ♦ red and yellow curves are mirror images of each other.
   ♦ magenta line is the mirror line.

4. y = e^x is the inverse of y = log_e x and vice versa.
If we combine them as a composite function, we will get an identity function. This can be demonstrated in (iii) steps:
(i) Let f(x) = e^x and g(x) = log_e x.
• Then f(g(x)) = f(log_e x) = $\rm{e^{\log_e x}}$
• Let $\rm{e^{\log_e x}}$ = u
⇒ log_e u = log_e x
⇒ u = x
⇒ f(g(x)) = u = x
(ii) Similarly, g(f(x)) = g(e^x) = log_e (e^x).
• Let log_e (e^x) = v
⇒ e^x = e^v.
⇒ v = x
⇒ g(f(x)) = v = x
(iii) From (i) and (ii), we get: f(g(x)) = g(f(x)) = x
• That means, f(x) is the inverse of g(x) and vice versa.
• In other words, y = e^x is the inverse of y = log_e x and vice versa.
• This true for all acceptable values of base b. We can write:
   ♦ y = 10^x is the inverse of y = log₁₀ x and vice versa.
   ♦ y = 2^x is the inverse of y = log₂ x and vice versa.
   ♦ y = e^x is the inverse of y = log_e x and vice versa.
   ♦ etc.,

Based on this information, let us see a solved example:

Solved example 21.42
Is it true that $\rm{x = e^{\log x}}$ for all real x?
Solution:
1. Consider $\rm{x = e^{\log x}}$.
• It is a composite of two functions:
    ♦ $\rm{y = e^x}$
    ♦ $\rm{y = \log_e x}$
(Recall that, in this chapter, we write $\rm{\log_e x}$ simply as $\rm{\log x}$)
2. Each of the two functions is inverse of the other.
• So whatever is the input, the output will be same as input.
3. Now suppose that, the input x is a −ve number.
• Then the "$\rm{\log x}$" portion of the composite function will not be able to process the input x. This is because, logarithm of −ve numbers does not exist.
4. Therefore, the equation $\rm{x = e^{\log x}}$ is true only when input x is +ve.

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In a logarithm function, if e is taken as the base, then we denote it as ln. This can be explained in 4 steps:
1. Recall that, in the case of the exponential function y = b^x, if the base is 10, we call it common exponential function.
• In a similar way, in the case of the logarithmic function y = log_b x, if the base is 10, we call it common logarithmic function.
2. Recall that, in the case of the exponential function y = b^x, if the base is e, we call it natural exponential function.
• In a similar way, in the case of the logarithmic function y = log_b x, if the base is e, we call it natural logarithmic function.
3. So the natural logarithmic function is: y = log_e x
• It can be written in a short form as: y = ln x.
4. In this chapter, if the base is not specified, it means natural logarithm.
• So, in this chapter, if we see y = log 5, then it means, y = the natural logarithm of 5.

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We have completed a basic discussion on logarithmic functions. In the next section, we will see properties of logarithmic functions.

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