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Question: Find the domains of definition of the following functions. \(f(x) = {\log _{2x - 5}}({x^2} - 3x - ...

Find the domains of definition of the following functions.
f(x)=log⁡2x−5(x2−3x−10).f(x) = {\log _{2x - 5}}({x^2} - 3x - 10).

Explanation

Solution

The domain of a function is the set into which all of the inputs of the function is constrained to follow. It is the set X in the notation,f:x→yf:x \to y. To solve the given function, we can use the formula of the logarithmic function as given below:
log⁡ba=log⁡ealog⁡eb{\log _b}a = \dfrac{{{{\log }_e}a}}{{{{\log }_e}b}}
and for solving quadratic equations, (x2−3x−10)\left( {{x^2} - 3x - 10} \right)we use splitting the equation by middle term.

Complete step by step solution:
f(x)=log⁡2x−5(x2−3x−10) =log⁡e(x2−3x−10)log⁡e(2x−5)−−−−(i)  f(x) = {\log _{2x - 5}}({x^2} - 3x - 10) \\\ = \dfrac{{{{\log }_e}({x^2} - 3x - 10)}}{{{{\log }_e}(2x - 5)}} - - - - (i) \\\
The equation can be factorized by splitting the middle terms as:
x2−3x−10=x2+2x−5x−10 =x(x+2)−5(x+2) =(x+2)(x−5)−−−−(ii)  {x^2} - 3x - 10 = {x^2} + 2x - 5x - 10 \\\ = x(x + 2) - 5(x + 2) \\\ = (x + 2)(x - 5) - - - - (ii) \\\
Substituting equation (ii) in equation (i) we get,
f(x)=log⁡e(x2−3x−10)log⁡e(2x−5) =log⁡(x+2)(x−5)log⁡(2x−5)−−−−(iii)  f(x) = \dfrac{{{{\log }_e}({x^2} - 3x - 10)}}{{{{\log }_e}(2x - 5)}} \\\ = \dfrac{{\log (x + 2)(x - 5)}}{{\log (2x - 5)}} - - - - (iii) \\\
Now, as the denominator of f(x) cannot be less than or even equals to zero so,
2x−5>0 2x>5 x>52−−−−(iv)  2x - 5 > 0 \\\ 2x > 5 \\\ x > \dfrac{5}{2} - - - - (iv) \\\
Also, the terms associated with the logarithmic function cannont be negative (less than zero) so,
x+2>0 and x−5>0 x>−2 and x>5 x>5−−−−(v)  x + 2 > 0{\text{ and }}x - 5 > 0 \\\ x > - 2{\text{ and }}x > 5 \\\ x > 5 - - - - (v) \\\
Also,
x+2<0 and x−5<0 x<\-2 and x<5 x<\-2−−−−(vi)  x + 2 < 0{\text{ and }}x - 5 < 0 \\\ x < \- 2{\text{ and }}x < 5 \\\ x < \- 2 - - - - (vi) \\\
By equation (v) and (vi), we can say that:
x∈(−∞,−2)∪(5,∞)−−−−(vii)x \in ( - \infty , - 2) \cup (5,\infty ) - - - - (vii)
Now from equation (iv) and (vii):
x∈(x>52)∩((−∞,−2)∪(5,∞)) x∈(52,∞)  x \in \left( {x > \dfrac{5}{2}} \right) \cap \left( {( - \infty , - 2) \cup (5,\infty )} \right) \\\ x \in \left( {\dfrac{5}{2},\infty } \right) \\\
From (i) and (ii) we get the domain of the function (52,∞)\left( {\dfrac{5}{2},\infty } \right)

Note: Students must have knowledge of inequalities of a function and logarithmic formula so that the student will be able to solve the question properly. We can also find the domain of this function with the graphical method by identifying the nature of the function as we know that logarithmic functions always give positive value.