The domain of f(x) = log_(x+1)(x-2) / (e^(2log_e x) - (2x+3)), x ∈ R is (1) R - {1, -3} (2) (2, ∞) -

Q: Wait, why do we reject the negative root when solving $x^2 - 2x - 3 \neq 0$?

Great catch! When we factor $x^2 - 2x - 3$ into $(x-3)(x+1)$, we get potential roots at $x=3$ and $x=-1$. While $x=-1$ is indeed a root of the denominator, it is already excluded by the condition $x+1 > 0$ (the base of the logarithm must be positive). So, while $x=-1$ is a 'forbidden' value, it's already covered by our earlier constraints. We focus on $x=3$ as the primary exclusion in our final domain.

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm! However, this is a pure mathematical domain problem. In mathematics, 'domain' refers to the set of all possible input values ($x$) for which the function is defined. We aren't looking for energy; we are looking for the 'allowable' values of $x$ that keep the function from becoming undefined (like dividing by zero or taking the log of a negative number).

Q: Is it really necessary to check $x \neq 0$ for the base of the log?

Absolutely! The base of a logarithm, $b$, must satisfy $b > 0$ and $b \neq 1$. Here, our base is $x+1$. If $x+1 = 1$, then $x=0$. Since $1$ raised to any power is always $1$, it cannot serve as a base for a logarithm (it would make the function constant or undefined). Thus, $x=0$ must be excluded from the domain.

Question

The domain of $f(x) = \frac{\log_{(x+1)}(x-2)}{e^{2\log_e x} - (2x+3)}, x \in \mathbb{R}$ is

Options:

(A) $\mathbb{R} - \{1, -3\}$

(B) $(2, \infty) - \{3\}$

(C) $(-1, \infty) - \{3\}$

(D) $\mathbb{R} - \{3\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The function is defined as:

$$f(x) = \frac{\log_{(x+1)}(x-2)}{e^{2\log_e x} - (2x+3)}$$

To find the domain, we must identify all values of $x$ for which the expression is well-defined. This requires satisfying the constraints of the logarithmic numerator, the logarithmic term in the denominator, and ensuring the denominator itself is non-zero.

### Phase 1: The Numerator's Rules

The numerator contains the term $\log_{(x+1)}(x-2)$. Logarithms are subject to two strict conditions:

1. **The Argument Rule:** The argument must be strictly positive, so $x - 2 > 0$, which implies $x > 2$.

2. **The Base Rule:** The base must be strictly positive and not equal to $1$. Thus, $x + 1 > 0$ (or $x > -1$) and $x + 1 
eq 1$ (or $x 
eq 0$).

Combining these, the numerator is defined for $x > 2$. Note that any $x > 2$ automatically satisfies $x > -1$ and $x 
eq 0$.

### Phase 2: The Denominator's Secrets

The denominator is $e^{2\log_e x} - (2x+3)$. We simplify the exponential term using logarithmic properties:

$$e^{2\log_e x} = e^{\log_e (x^2)} = x^2$$

Thus, the denominator simplifies to $x^2 - 2x - 3$. We must satisfy two additional conditions here:

1. **Log Existence:** The term $\log_e x$ requires $x > 0$.

2. **Non-Zero Denominator:** The denominator cannot be zero, so $x^2 - 2x - 3 
eq 0$. Factoring the quadratic, we get:

$$(x - 3)(x + 1) 
eq 0$$

This implies $x 
eq 3$ and $x 
eq -1$.

### Phase 3: The Final Intersection

We now aggregate all constraints derived from the function:

*   From the numerator: $x > 2$
*   From the base: $x > -1$ and $x 
eq 0$
*   From the denominator log: $x > 0$
*   From the denominator non-zero: $x 
eq 3$ and $x 
eq -1$

The condition $x > 2$ is the most restrictive, as it inherently satisfies $x > 0$ and $x > -1$. We must also ensure that $x 
eq 3$ is respected within the interval $(2, \infty)$.

The final domain of the function is:

**$$x \in (2, 3) \cup (3, \infty)$$**

Animated Solution for Mathematics - Functions: The domain of $f (x) = \frac{l o g _{(x + 1)} ( x - 2 )}{e ^{2 l o g_{e} x} - ( 2 x + 3 )}, x \in R$ is

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Visualized Solution

Understanding the Function Structure

Numerator Constraint: Argument of Log

Numerator Constraint: Base of Log

Numerator Constraint: Base $\neq =$ 1

Denominator Constraint: Log Existence

Simplifying the Denominator

Solving for Denominator Zeros

Final Intersection and Domain

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The domain of f(x)=e2loge​x−(2x+3)log(x+1)​(x−2)​,x∈R is