21.15 - Derivatives of Exponential and Logarithmic Functions

In the previous section, we completed a discussion on exponential and logarithmic functions. In this section, we will see the derivatives of those two functions.

First we will see the derivative of exponential function.
• If f(x) = e^x, then it’s derivative is the same e^x.
• That is., if f(x) = e^x, then f'(x) = e^x.
    ♦ In other words, $\rm{\text{if}~y = e^x,~\text{then},~\frac{dy}{dx} = e^x}$
• We will see the proof in higher classes. At present, we will see a simple application of this derivative.  It can be written in 5 steps:

1. The red curve in fig.21.20 below shows the graph of f(x) = e^x.

Fig.21.20

2. Mark any convenient point on the curve. Let us mark the point with x-coordinate 1.5.
• Since the x-coordinate is 1.5,
y-coordinate = e^1.5 = 4.4816
• We will use a single decimal place and write:
e^1.5 = 4.5.
• So the coordinates are (1.5,4.5). We will name this point as P
 

3. Next, we want the derivative of f(x) at x = 1.5
• That is, we want f'(1.5).
• We wrote that, f'(x) is the same e^x.
So f'(1.5) = e^1.5 = 4.5

4. We know that, f'(1.5) will be the slope of the tangent at x = 1.5
• Let us draw a line through P, at a slope of f'(1.5).
We have a point P(1.5,4.5) and slope 4.5. So the equation of this line will be:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y-y_1}    & {~=~}    &{\text{slope} \times (x - x_1)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y-4.5}    & {~=~}    &{4.5 \times (x – 1.5)}    \\
{~\color{magenta}    3    }    &{\implies}    &{y-4.5}    & {~=~}    &{4.5 x \,–\, 6.75}    \\
{~\color{magenta}    4    }    &{\implies}    &{y}    & {~=~}    &{4.5 x \,–\, 2.25}    \\
\end{array}$                           

• Let us plot this line. It is shown in green color in fig.21.20 above.
(note that, the y-intercept of the green line in the fig.21.20 is −2.25)
• We see that, the green line is the tangent at P.   

5. Let us write a summary:
(i) We calculated e^1.5, which is the e^x at P(1.5,4.5).
(ii) We drew a line through P(1.5,4.5) at a slope equal to e^1.5.
(iii) That line happens to be the tangent at P. So e ^1.5 is the derivative at P
(iv) Therefore, the general form of the derivative is e^x.

Note that, the above demonstration is not a proof. We will see the actual proof in higher classes.

---

Now we will see the derivative of log function.
• If f(x) = log x, then it’s derivative is $\rm{\frac{1}{x}}$.
• That is., if f(x) = log x, then f'(x) = $\rm{\mathbf{{\frac{1}{x}}}}$.
    ♦ In other words, $\rm{\text{if}~y = \log x,~\text{then},~\frac{dy}{dx} = \frac{1}{x}}$
(Recall that, in this chapter, when we write log x, it means, base is e)
• We will see the proof in higher classes. At present, we will see a simple application of this derivative.  It can be written in 5 steps:

1. The red curve in fig.21.21 below shows the graph of f(x) = log x.

Fig.21.21

2. Mark any convenient point on the curve. Let us mark the point with x-coordinate 2.0
• Since the x-coordinate is 2.0,
y-coordinate = log 2 = 0.6931
• We will use a single decimal place and write:
log = 0.7
• So the coordinates are (2,0.7). We will name this point as P
 

3. Next, we want the derivative of f(x) at x = 2
• That is, we want f'(2).
• We wrote that, f'(x) is $\frac{1}{x}$.
So f'(2) = $\rm{\frac{1}{2}}$ = 0.5

4. We know that, f'(2) will be the slope of the tangent at x = 2
• Let us draw a line through P, at a slope of f'(2).
We have a point P(2,0.7) and slope 0.5. So the equation of this line will be:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y-y_1}    & {~=~}    &{\text{slope} \times (x - x_1)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y-0.7}    & {~=~}    &{0.5 \times (x – 2)}    \\
{~\color{magenta}    3    }    &{\implies}    &{y-0.7}    & {~=~}    &{0.5 x \,–\, 1}    \\
{~\color{magenta}    4    }    &{\implies}    &{y}    & {~=~}    &{0.5 x \,–\, 0.3}    \\
\end{array}$                           

• Let us plot this line. It is shown in green color in fig.21.21 above.
(note that, the y-intercept of the green line in the fig.21.20 is −0.3)
• We see that, the green line is the tangent at P.   

5. Let us write a summary:
(i) We calculated $\frac{1}{x}$  at P(2,0.7).
(ii) We drew a line through P(2,0.7) at a slope equal to $\frac{1}{2}$.
(iii) That line happens to be the tangent at P. So $\frac{1}{x}$ is the derivative at P
(iv) Therefore, the general form of the derivative is $\frac{1}{x}$.

Note that, the above demonstration is not a proof. We will see the actual proof in higher classes.

---

Now we will see some solved examples:

Solved example 21.50
Differentiate the following w.r.t to x:
(i) e^−x    (ii) sin(log x), x>0    (iii) cos⁻¹(e^x)    (iv) e^{cos x}.
Solution:
Part (i):

◼ Remarks:
• 3(Magenta color): Here we apply chain rule.

Alternate method:

◼ Remarks:
• 2(Magenta color): Here we take logarithm on both sides.
• 4(Magenta color): Here we apply chain rule.
• 5(Magenta color): Here we apply the fact that, log_ee = 1.

Part (ii):

◼ Remarks:
• 3(Magenta color): Here we apply chain rule.

Part (iii):

◼ Remarks:
• 3(Magenta color): Here we apply chain rule. 

Part (iv):

◼ Remarks:
• 3(Magenta color): Here we apply chain rule.

---

Link to a few more solved examples is given below:

Exercise 21.4

---

In the next section, we will see logarithmic differentiation.

Previous

Contents


Next

Copyright©2024 Higher secondary mathematics.blogspot.com

21.14 - Solution of Exponential and Logarithmic Functions

In the previous section, we saw properties of logarithms. In this section, we will see some solved examples which demonstrate the process of solving exponential and logarithmic equations.

Solved example 21.46
Solve the equation: log(6x) − log(4-x) = log 3
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log (6x) - \log (4-x)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log \left(\frac{6x}{4-x} \right)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{6x}{4-x}}    & {~=~}    &{3}    \\
{~\color{magenta}    4    }    &{\implies}    &{6x}    & {~=~}    &{12 – 3x}    \\
{~\color{magenta}    5    }    &{\implies}    &{9x}    & {~=~}    &{12}    \\
{~\color{magenta}    6    }    &{\implies}    &{x}    & {~=~}    &{\frac{4}{3}}    \\
\end{array}$                           

Check:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log (6(4/3)) - \log (4- 4/3)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log (8) - \log (8/3)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{8}{8/3}}    & {~=~}    &{3}    \\
{~\color{magenta}    4    }    &{\implies}    &{3}    & {~=~}    &{3}    \\
\end{array}$

Solved example 21.47
Solve the equation: ln(4 −3x) − ln(7x) = ln(11)
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\ln(4-3x) - \ln(7x)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\ln\left[\frac{4-3x}{7x} \right]}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{4-3x}{7x}}    & {~=~}    &{11}    \\
{~\color{magenta}    4    }    &{\implies}    &{4 – 3x}    & {~=~}    &{77x}    \\
{~\color{magenta}    5    }    &{\implies}    &{80x}    & {~=~}    &{4}    \\
{~\color{magenta}    6    }    &{\implies}    &{x}    & {~=~}    &{\frac{1}{20}}    \\
\end{array}$                            
 

Check:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\ln(4-3x) - \ln(7x)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\ln(4-3(1/20)) - \ln(7(1/20))}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\ln(77/20) - \ln(7/20)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    4    }    &{\implies}    &{\ln(77) - \ln(20) - \ln(7) + \ln(20)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    5    }    &{\implies}    &{\ln(77) - \ln(7)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    6    }    &{\implies}    &{\ln(77/7)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    7    }    &{\implies}    &{\ln(11)}    & {~=~}    &{\ln(11)}    \\
\end{array}$

Solved example 21.48
Solve the equation: log₈ (4x + 1) = −1
Solution:

Check:

Solved example 21.49
Solve the equation: 2e^3y+8 − 11e^5−10y = 0
Solution:

Check:

---

We have seen the process of solving exponential and logarithmic equations. The reader is advised to try a large number of practice problems in this category. In the next section, we will see derivatives of exponential and logarithmic functions.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

21.13 - Properties of Logarithm

In the previous section, we saw some basic details about logarithmic functions. In this section, we will see some of their properties.

Property I: $\rm{\log_a p \,=\, \frac{\log_b p}{\log_b a}}$
   ♦ On the L.H.S, we have log of p to the base a.
   ♦ On the R.H.S, we have log of p to the base b.
• So this property gives us the method to connect two logs of the same number when the bases are different.
• This property is also known as change of base rule.
• We can write the proof in 5 steps:

1. Let $\rm{\log_a p = \alpha}$
Then $\rm{a^{\alpha} = p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b a = \gamma}$
Then $\rm{b^{\gamma} = a}$
4. Substituting (3) in (1), we get:
$\rm{\left(b^{\gamma} \right)^{\alpha} = p}$
⇒ $\rm{b^{\gamma \alpha} = p}$
5. Substituting (4) in (2), we get:
$\rm{b^{\beta} = p = b^{\gamma \alpha}}$
⇒ $\rm{\beta = \gamma \alpha}$
⇒ $\rm{\alpha = \frac{\beta}{\gamma}}$
⇒ $\rm{\log_a p = \frac{\log_b p}{\log_b a}}$

Property II: $\rm{\log_b pq \,=\, \log_b p + \log_b q}$
   ♦ On the L.H.S, we have log of a product pq.
   ♦ On the R.H.S, we have addition of two individual logs.
• So this property helps us to simplify logarithmic equations when products are involved.
• We can write the proof in 4 steps:
1. Let $\rm{\log_b pq = \alpha}$
Then $\rm{b^{\alpha} = pq}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b q = \gamma}$
Then $\rm{b^{\gamma} = q}$
4. Substituting (2) and (3) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\gamma}$
⇒ $\rm{b^\alpha = b^{\beta + \gamma}}$
⇒ $\rm{\alpha = \beta + \gamma}$
⇒ $\rm{\log_b pq = \log_b p + \log_b q}$

Property III: $\rm{\log_b \left(\frac{p}{q} \right) \,=\, \log_b p - \log_b q}$
   ♦ On the L.H.S, we have log of a ratio $\frac{p}{q}$.
   ♦ On the R.H.S, we have subtraction of one individual log from another.
• So this property helps us to simplify logarithmic equations when ratios are involved.
• We can write the proof in 4 steps:
1. Let $\rm{\log_b \left(\frac{p}{q} \right) = \alpha}$
Then $\rm{b^{\alpha} = \frac{p}{q}}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b q = \gamma}$
Then $\rm{b^{\gamma} = q}$
4. Substituting (2) and (3) in (1), we get:
$\rm{b^\alpha = \frac{b^\beta}{b^\gamma}}$
⇒ $\rm{b^\alpha = b^{\beta - \gamma}}$
⇒ $\rm{\alpha = \beta - \gamma}$
⇒ $\rm{\log_b pq = \log_b p - \log_b q}$

Property IV: $\rm{\log_b p^2 \,=\, 2 \log_b p}$
   ♦ On the L.H.S, we have log of a square.
   ♦ On the R.H.S, we have individual log multiplied by 2.
• So this property helps us to simplify logarithmic equations when squares are involved.
• We can write the proof in 3 steps:
1. Let $\rm{\log_b (p \times p) = \alpha}$
Then $\rm{b^{\alpha} = p \times p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Substituting (2) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\beta}$
⇒ $\rm{b^\alpha = b^{\beta + \beta}}$
⇒ $\rm{\alpha = \beta + \beta}$
⇒ $\rm{\log_b (p \times p) = \log_b p + \log_b p}$
⇒ $\rm{\log_b p^2 = 2 \log_b p}$

Property V: $\rm{\log_b p^3 \,=\, 3 \log_b p}$
   ♦ On the L.H.S, we have log of a cube.
   ♦ On the R.H.S, we have individual log multiplied by 3.
• So this property helps us to simplify logarithmic equations when cubes are involved.
• We can write the proof in 3 steps:
1. Let $\rm{\log_b (p \times p \times p) = \alpha}$
Then $\rm{b^{\alpha} = p \times p \times p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Substituting (2) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\beta \times b^\beta}$
⇒ $\rm{b^\alpha = b^{\beta + \beta + \beta}}$
⇒ $\rm{\alpha = \beta + \beta + \beta}$
⇒ $\rm{\log_b (p \times p \times p) = \log_b p + \log_b p + \log_b p}$
⇒ $\rm{\log_b p^3 = 3 \log_b p}$

Property VI: $\rm{\log_b p^n \,=\, n \log_b p}$
   ♦ On the L.H.S, we have log of a n^th power.
   ♦ On the R.H.S, we have individual log multiplied by n.
• So this property helps us to simplify logarithmic equations when n^th power is involved.
• The proof can be written using the principles of mathematical induction.

Property VII: $\rm{\log_p p \,=\, 1}$
   ♦ On the L.H.S, we have same number and base.
   ♦ On the R.H.S, we have 1.
We can write the proof as follows:
• Let $\rm{\log_p p = \alpha}$
• Then $\rm{p^{\alpha} = p}$
⇒ $\rm{p^{\alpha} = p^1}$
⇒ $\rm{\alpha = 1}$

Property VIII: $\rm{\log_b p \,=\, \frac{1}{\log_p b}}$
   ♦ On the L.H.S, we have base b and number p.
   ♦ On the R.H.S, we have base p and number b.
   ♦ So base and number are interchanged.
We can write the proof as follows:
• Applying the change of base rule we get:
$\rm{\log_b p \,=\, \frac{\log_p p}{\log_p b} \,=\, \frac{1}{\log_p b}}$ 

Property IX: $\rm{\log_p {p^x} \,=\, x}$
We can write the proof as follows:
• Let $\rm{\log_p p^x = \alpha}$
• Then $\rm{p^{\alpha} = p^x}$
⇒ $\rm{\alpha = x}$

• Note that, this property is used when we first begin to learn about logarithms. For example:
$\rm{\log_5 625 = \log_5 5^4 = 4}$

---

Now we will see some solved examples

Solved example 21.43
Express as a single logarithm:
$\rm{2 \log x - 5 \log y + 3 \log z}$
Solution:
1. 2 log x = log x².
2. −5 log y = −log y⁵.
3. 3 log z = log z³.
4. log x² − log y⁵ + log z³ = $\rm{\frac{x^2 z^3}{y^5}}$

Solved example 21.44
Expand using the properties of logarithm:
$\rm{\log_5 \sqrt[3]{x}}$
Solution:
$\rm{\log_5 \sqrt[3]{x}}$
= $\rm{\log_5 \left(x^{\frac{1}{3}} \right)}$ 
= $\rm{\frac{1}{3} \log_5 x}$

Solved example 21.45
Evaluate log₅ 28
Solution:
1. We could find the solution easily if it is "25" instead of "28".
2. We cannot directly use the calculator because base is 5.
3. So we will apply the change of base rule. We get:
$\rm{\log_5 28 \,=\,\frac{\log_{10} 28}{\log_{10} 5}\,=\,\frac{1.44715}{0.69897}\,=\,2.07040}$

---

In the next section, we will see exponential and logarithmic equations.

Previous

Contents


Next

Copyright©2024 Higher secondary mathematics.blogspot.com

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.

---

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

---

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.

---

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.

---

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.

---

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.

---

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

Previous

Contents


Next

Copyright©2024 Higher secondary mathematics.blogspot.com