Transcendental Function - Explanation, Equation and Examples

Transcendental function is a category of functions that cannot be expressed as finite polynomials, algebraic functions, or solutions of algebraic equations with rational coefficients. They are crucial in advanced mathematics and have applications across various scientific fields. These are non-algebraic functions.

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In this article, we will cover the concepts of the transcendental function especially exponential and logarithmic functions. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of seven questions have been asked on this concept, including two in 2019, and five in 2023.

Define Transcendental Functions

The transcendental function can be defined as a function that is not algebraic and cannot be expressed in terms of a finite sequence of algebraic operations such as sin x.

The most familiar transcendental functions examples are the exponential functions, logarithmic functions, trigonometric functions, hyperbolic functions, and inverse of all these functions. The less familiar transcendental functions examples are Gamma, Elliptic, and Zeta functions.

Types of Transcendental function

Exponential Function: function $f(x)$ such that $f(x)=a^x$ is known as an exponential function.
base: $a>0,a\ne 1$
domain : $x\in \mathbb{R}$
range : $\mathrm{f}(\mathrm{x})>0$

Property : If $a^x=a^y$,then $x=y$

Logarithmic function: function $f(x)$ such that $f(x)={\log }_a(x)$ is called logarithmic function
base: $a>0,a\ne 1$
domain : $\mathrm{x}>0$
range $:f(x)\in \mathbb{R}$

If a > 1 If 0 < a < 1

Properties of Logarithmic Function

1. ${\log }_e(ab)={\log }_{e}a+{\log }_{e}b$
2. ${\log }_e\left(\frac{a}{b}\right)={\log }_{e}a-{\log }_{e}b$
3. ${\log }_e{a}^{\mathrm{m}}=m{\log }_{e}a$
4. ${\log }_{a}a=1$
5. ${\log }_{{\mathrm{b}}^{\mathrm{m}}}\mathrm{a}=\frac{1}{\mathrm{m}}{\log }_{\mathrm{b}}\mathrm{a}$
6. ${\log }_{b}a=\frac{1}{{\log }_{\mathrm{a}}\mathrm{b}}$
7. ${\log }_{b}a=\frac{{\log }_{m}a}{{\log }_{m}b}$
8. ${\mathrm{a}}^{{\log }_{a}m}=\mathrm{m}$
9. ${\mathrm{a}}^{{\log }_c\mathrm{b}}={\mathrm{b}}^{{\log }_c\mathrm{a}}$
10. ${\log }_{\mathrm{m}}\mathrm{a}=\mathrm{b}\Rightarrow \mathrm{a}={\mathrm{m}}^{\mathrm{b}}$

Summary

Transcendental functions are essential tools in mathematics and science, extending beyond simple algebraic functions to describe more complex and varied phenomena. Their unique properties and applications make them indispensable in theoretical and applied contexts.

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Solved Examples Based On the Transcendental Functions:

Example 1: What is the range of the function $f(x)=\left|a^x\right|$ ?
1) $[1,\mathrm{\infty })$
2) $R$
3) $R^{+}$
4) $[0,\mathrm{\infty })$

Solution:

$\begin{array}{rl}& a^x>0\mathrm{\forall }x\equiv R\\ & \text{Thus}\left|a^x\right|=a^x\\ & \text{Hence}\left|a^x\right|>0\mathrm{\forall }x\equiv R\\ & \text{i.e}\left|a^x\right|\in R^{+}\end{array}$
Hence, the answer is the option 3.

Example 2: The number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$ is:
Solution:

$\begin{array}{rl}& 5+|2^x-1|=2^x(2^x-2)\\ & 5+|2^x-1|=2^{2x}-2\cdot 2^x\end{array}$

$\begin{array}{rl}& \text{Case 1:}x\ge 0\\ & =>5+2^x-1=2^{2x}-2\cdot 2^x\\ & =>2^{2x}-2\cdot 2^x-2^x-4=0\\ & =>(2^x-4)(2^x+1)=0\\ & =>2^x-4=0\\ & \Rightarrow x=2\end{array}$

Case 2: $x<0$

$\begin{array}{rl}& 5+1-2^x=2^{2x}-2\cdot 2^x\\ & =>6=2^{2x}-2^x\\ & \text{LHS}=+\text{ve}\mathrm{\&}\text{RHS}=-\text{ve (no solution)}\\ & \therefore \text{no. of solution}=1\end{array}$

Hence, the answer is 1.
Example 3: Which of the following is the domain of $f(x)={\log }_2(x+4)$
1) $(0,\mathrm{\infty })$
2) $(-2,\mathrm{\infty })$
3) $(-4,\mathrm{\infty })$

4) None of these

Solution:$\begin{array}{rl}& \ln {\log }_2(x+4);x+4>0\\ & \text{i.e}x>-4\end{array}$

Hence, the answer is the option 3.
Example 4: If $f(x)={\log }_c\left(\frac{1-x}{1+x}\right),|x|<1,\underset{\text{then}}{}f\left(\frac{2x}{1+x^2}\right)$ is equal to
1) $2f\left(x^2\right)$
2) $2f(x)$
3) $-2f(x)$
4) $(f(x){)}^2$

Solution:

$\begin{array}{rl}& f(x)={\log }_e\left(\frac{1-x}{1+x}\right),|x|<1\\ & f\left(\frac{2x}{1+x^2}\right)=?\\ & f\left(\frac{2x}{1+x^2}\right)={\log }_e\left(\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right)\\ & ={\log }_e\left(\frac{\frac{1+x^2-2x}{1+x^2}}{\frac{1+x^2+2x}{1+x^2}}\right)\\ & ={\log }_e\frac{x^2-2x+1}{x^2+2x+1}={\log }_e\left(\frac{(1-x{)}^2}{(x+1{)}^2}\right)\\ & ={\log }_e{\left(\frac{1-x}{x+1}\right)}^2\end{array}$

$\begin{array}{rl}& =2{\log }_e\left(\frac{1-x}{x+1}\right),|x|<1\\ & =2f(x)\end{array}$

Hence, the answer is the option 2.
Example 5: The number of solutions of the equation ${\log }_4(x-1)={\log }_2(x-3)$ is
Solution:

$\begin{array}{rl}& {\log }_4(x-1)={\log }_2(x-3)\\ & \Rightarrow \frac{1}{2}{\log }_2(x-1)={\log }_2(x-3)\\ & \Rightarrow {\log }_2(x-1{)}^{1/2}={\log }_2(x-3)\\ & \Rightarrow \sqrt{x-1}=(x-3)\\ & \Rightarrow x-1=x^2+9-6x\\ & \Rightarrow x^2-7x+10=0\\ & \Rightarrow (x-2)(x-5)=0\\ & \Rightarrow x=2,5\end{array}$

but $x$ can't be 2 , because in

${\log }_2(x-3)\to x>3$

So, there is only one solution
Hence, the answer is 1 .