3.10 Domain and range of exponential and logarithmic function  (Page 26/2)

Problem : Find the domain of the function given by (Be aware that "x" appears as base of given logrithmic function):

$$f\left(x\right)={\log }_{x}2$$

Solution : By definition of logarithmic function, we know that base of logarithmic function is a positive number excluding x =1.

$$x>\mathrm{0,}x\ne 1$$

Hence, domain of the given function is :

Domain of the function

$$\text{Domain}=\left(\mathrm{0,}\infty \right)-\left\{1\right\}$$

or,

$$\text{Domain}=\left(\mathrm{0,1}\right)\cup \left\{\mathrm{1,}\infty \right\}$$

Problem : Find the domain of the function given by :

$$f\left(x\right)={\log }_{10}\frac{x^2-5x+6}{x^2+5x+9}$$

Solution : The argument (input to the function) of logarithmic function is a rational function. We need to find values of “x” such that the argument of the function evaluates to a positive number. Hence,

$$\Rightarrow \frac{x^2-5x+6}{x^2+5x+9}>0$$

In this case, we can not apply sign scheme for the rational function as a whole. Reason is that the quadratic equation in the denominator has no real roots and as such can not be factorized in linear factors. We see that discreminant,"D", of the quadratic equation in the denominator, is negative :

$$\Rightarrow D=b^2-4ac=5^2-4X1X9=25-36=-11$$

The quadratic expression in denominator is positive for all value of x as coefficient of squared term is positive. Clearly, sign of rational function is same as that of quadratic expression in the numerator. The coefficient of squared term of the numerator “ $x^2$ ”, is positive for all values of “x”. The quadratic expression in the numerator evaluates to positive for intervals beyond root values. The roots of the corresponding equal equation is :

$$\Rightarrow x^2-2x-3x+6=0\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\Rightarrow \left(x-2\right)\left(x-3\right)=0$$

Domain of the function

$$x<2\text{or}x>3$$

$$\text{Domain}=\left(-\infty ,2\right)\cup \left(\mathrm{3,}\infty \right)$$