If the domain of the function f(x) = log(10x^2-17x+7) (18x^2 - 11x + 1) is (-∞, a) ∪ (b, c) ∪ (d, ∞)

Q: Wait, why do we reject the negative root here? Or rather, why do we have to exclude $x = 1/2$ and $x = 6/5$?

Great question! The condition $B(x) \neq 1$ is not just a suggestion; it is a mathematical necessity. If the base of a logarithm is $1$, the expression $\log_{1} A$ is undefined because there is no power to which you can raise $1$ to get $A$ (unless $A=1$, which is still indeterminate). By solving $10x^2 - 17x + 7 = 1$, we find the exact points where the base becomes $1$. We must exclude these points from our domain, even if they satisfy the other conditions.

Q: Why can't we just use energy conservation or some other shortcut? Why is the domain so complicated?

In mathematics, especially with functions, the domain is the 'foundation' of the building. If the foundation is flawed, the entire structure collapses. We cannot evaluate the function at points where it is not defined. The complexity arises because we are dealing with multiple constraints simultaneously. Think of it as a filter: each condition filters out invalid values of $x$. We are simply finding the values that survive all three filters.

Q: Is there a trick to avoid the long calculation at the end?

The trick is to be smart with your algebra! Instead of finding a common denominator for the sum $a + b + c + d + e$, notice that the multiplier is $90$. Since $90$ is a multiple of all the denominators ($9, 2, 10, 1, 5$), distributing the $90$ inside the parenthesis turns all those fractions into simple integers. Always look for these algebraic 'shortcuts' before diving into tedious arithmetic.

Question

If the domain of the function $f(x) = \log_{(10x^2-17x+7)} (18x^2 - 11x + 1)$ is $(-\infty, a) \cup (b, c) \cup (d, \infty) - \{e\}$, then $90(a + b + c + d + e)$ equals:

Options:

(A) 170

(B) 307

(C) 316

(D) 177

Sigma Percentile · Accepted Answer

### The Logarithmic Gatekeeper: Understanding Domain Constraints

To find the domain of the function $f(x) = \log_{(10x^2-17x+7)} (18x^2 - 11x + 1)$, we must satisfy three non-negotiable conditions for a logarithm $\log_{B(x)} A(x)$:
1. The argument must be positive: $A(x) > 0$.
2. The base must be positive: $B(x) > 0$.
3. The base must not equal unity: $B(x) 
eq 1$.

### Phase 1: The Argument's Domain

We first solve the inequality for the argument:
$$18x^2 - 11x + 1 > 0$$

Factorizing the quadratic expression:
$$18x^2 - 9x - 2x + 1 > 0 \Rightarrow 9x(2x - 1) - 1(2x - 1) > 0 \Rightarrow (9x - 1)(2x - 1) > 0$$

The critical points are $x = 1/9$ and $x = 1/2$. Since the inequality is strictly greater than zero, the solution is:
$$x \in \left(-\infty, \frac{1}{9}ight) \cup \left(\frac{1}{2}, \inftyight)$$

### Phase 2: The Base's Domain

Next, we ensure the base is positive:
$$10x^2 - 17x + 7 > 0$$

Factorizing the quadratic:
$$10x^2 - 10x - 7x + 7 > 0 \Rightarrow 10x(x - 1) - 7(x - 1) > 0 \Rightarrow (10x - 7)(x - 1) > 0$$

The critical points are $x = 7/10$ and $x = 1$. The solution for this condition is:
$$x \in \left(-\infty, \frac{7}{10}ight) \cup (1, \infty)$$

### Phase 3: The Hidden Constraint

We must also ensure the base is not equal to $1$:
$$10x^2 - 17x + 7 
eq 1 \Rightarrow 10x^2 - 17x + 6 
eq 0$$

Factorizing the expression:
$$(2x - 1)(5x - 6) 
eq 0$$

This implies that $x 
eq 1/2$ and $x 
eq 6/5$. These are the **forbidden points** that must be excluded from our final set.

### Phase 4: The Intersection

We now find the intersection of the conditions:
$$\left[ \left(-\infty, \frac{1}{9}ight) \cup \left(\frac{1}{2}, \inftyight) ight] \cap \left[ \left(-\infty, \frac{7}{10}ight) \cup (1, \infty) ight]$$

The overlapping intervals are:
$$\left(-\infty, \frac{1}{9}ight) \cup \left(\frac{1}{2}, \frac{7}{10}ight) \cup (1, \infty)$$

Applying the exclusions $x 
eq 1/2$ and $x 
eq 6/5$, we note that $1/2$ is already excluded by the open interval. However, $6/5$ must be explicitly removed. The final domain is:
$$\left(-\infty, \frac{1}{9}ight) \cup \left(\frac{1}{2}, \frac{7}{10}ight) \cup (1, \infty) \setminus \left\{ \frac{6}{5} ight\}$$

### Phase 5: The Final Payoff

We identify the values $a = 1/9$, $b = 1/2$, $c = 7/10$, $d = 1$, and $e = 6/5$. We calculate $90(a + b + c + d + e)$:

$$90 \left( \frac{1}{9} + \frac{1}{2} + \frac{7}{10} + 1 + \frac{6}{5} ight)$$

Distributing the $90$:
$$10 + 45 + 63 + 90 + 108 = 316$$

The final result is **316**.

Animated Solution for Mathematics - Functions: If the domain of the function $f (x) = lo g_{(10 x^{2} - 17 x + 7)} (18 x^{2} - 11 x + 1)$ is $(- \infty, a) \cup (b, c) \cup (d, \infty) - {e}$ , then $90 (a + b + c + d + e)$ equals:

Select Answer:

Visualized Solution

$f (x) = lo g_{B (x)} A (x)$ Domain Conditions

Condition 1: Argument $> 0$

Solving for Argument

Condition 2: Base $> 0$

Solving for Base

Condition 3: Base $\neq = 1$

Excluding Base $= 1$

Finding the Intersection

Final Domain

Identifying $a, b, c, d, e$

Final Calculation Setup

Distributing the $90$

Final Result

The Sigma Insight: Domain and Range of a Function

The Logarithmic Gatekeeper

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

If the domain of the function $f (x) = lo g_{7} (1 - lo g_{4} (x^{2} - 9 x + 18))$ is $(α, β) \cup (γ, δ)$ , then $α + β + γ + δ$ is equal to

Let the domain of the function $f (x) = lo g_{3} lo g_{5} (7 - lo g_{2} (x^{2} - 10 x + 85)) + sin^{- 1} (\frac{3 x - 7}{17 - x})$ be $(α, β]$ . Then $α + β$ is equal to :

If the domain of the function $sin^{- 1} (\frac{3 x - 22}{2 x - 19}) + lo g_{e} (\frac{3 x ^{2} - 8 x + 5}{x ^{2} - 3 x - 10})$ is $(α, β]$ , then $3 α + 10 β$ is equal to:

If domain of the function $lo g_{e} (\frac{6 x ^{2} + 5 x + 1}{2 x - 1}) + cos^{- 1} (\frac{2 x ^{2} - 3 x + 4}{3 x - 5})$ is $(α, β) \cup (γ, δ)$ , then $18 (α^{2} + β^{2} + γ^{2} + δ^{2})$ is equal to

If the domain of the function $f (x) = sin^{- 1} (\frac{5 - x}{3 + 2 x}) + \frac{1}{l o g _{e} ( 10 - x )}$ is $(- \infty, α] \cup [β, γ) - {δ}$ , then $6 (α + β + γ + δ)$ is equal to

If the domain of the function $f (x) = lo g_{e} (\frac{2 x + 3}{4 x ^{2} + x - 3}) + cos^{- 1} (\frac{2 x - 1}{x + 2})$ is $(α, β]$ , then the value of $5 β - 4 α$ is equal to

If the domain of the function $f (x) = lo g_{e} (\frac{2 x - 3}{5 + 4 x}) + sin^{- 1} (\frac{4 + 3 x}{2 - x})$ is $[α, β)$ , then $α^{2} + 4 β$ is equal to

If the domain of the function $f (x) = \frac{x ^{2} - 25}{( 4 - x ^{2} )} + lo g_{10} (x^{2} + 2 x - 15)$ is $(- \infty, α) \cup [β, \infty)$ , then $α^{2} + β^{3}$ is equal to :

Let the domain of the function $f (x) = cos^{- 1} (\frac{4 x + 5}{3 x - 7})$ be $[α, β]$ and the domain of $g (x) = lo g_{2} (2 - 6 lo g_{27} (2 x + 5))$ be $(γ, δ)$ . Then $∣7 (α + β) + 4 (γ + δ) ∣$ is equal to ________

Let the domains of the functions $f (x) = lo g_{4} lo g_{3} lo g_{7} (8 - lo g_{2} (x^{2} + 4 x + 5))$ and $g (x) = sin^{- 1} (\frac{7 x + 10}{x - 2})$ be $(α, β)$ and $[γ, δ]$ , respectively. Then $α^{2} + β^{2} + γ^{2} + δ^{2}$ is equal to :-

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=logB(x)​A(x) Domain Conditions

Condition 1: Argument .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } >0

Solving for Argument

Condition 2: Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } >0