The sum of all the real solutions of the equation log(x+3) (6x² + 28x + 30) = 5 - 2 log(6x+10) (x² +

Q: Why can't we just expand the logs immediately?

Expanding without factoring is like trying to solve a puzzle with your eyes closed. Factorization reveals the hidden structure—the common terms that allow for the elegant substitution. Without it, you are just moving algebraic clutter around.

Q: Do I really need to check the domain? It seems like extra work.

In the world of JEE Advanced, the domain is not 'extra work'; it is the foundation. Logarithms are only defined for positive bases (not equal to 1) and positive arguments. Skipping this check is the fastest way to lose marks on a 'trap' question. Always verify your roots against the original constraints.

Q: Why did the $6x$ terms cancel out?

That is the beauty of a well-crafted problem. When you set up the equation correctly, the algebraic structure is designed to simplify. If your terms aren't canceling, re-check your factorization of the quadratics. The cancellation is your signal that you are on the right path.

Question

The sum of all the real solutions of the equation $\log_{(x+3)} (6x^2 + 28x + 30) = 5 - 2 \log_{(6x+10)} (x^2 + 6x + 9)$ is equal to

Options:

(A) 2

(B) 1

(C) 0

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given equation is:

$$\log_{(x+3)} (6x^2 + 28x + 30) = 5 - 2 \log_{(6x+10)} (x^2 + 6x + 9)$$

In JEE Advanced, complexity is often a mask for elegance. Our mission is to peel back that mask by identifying the underlying structure.

### The Detective Work of Factorization

The first rule of engagement is to simplify the components. Let us factorize the quadratic expressions:

For $6x^2 + 28x + 30$, we factor out a $2$ to get $2(3x^2 + 14x + 15)$, which simplifies to:

$$2(3x + 5)(x + 3)$$

For $x^2 + 6x + 9$, we recognize the perfect square:

$$(x + 3)^2$$

The base $6x + 10$ is simply $2(3x + 5)$. The terms $(x + 3)$ and $(3x + 5)$ are the heartbeat of this equation.

### The Logarithmic Dance

We rewrite the equation using these factors:

$$\log_{(x+3)} [2(3x + 5)(x + 3)] = 5 - 2 \log_{2(3x+5)} (x + 3)^2$$

Using the product rule $\log_b (mn) = \log_b m + \log_b n$, the left side becomes:

$$\log_{(x+3)} [2(3x + 5)] + \log_{(x+3)} (x + 3) = \log_{(x+3)} [2(3x + 5)] + 1$$

On the right side, we use the power rule $\log_b (m^n) = n \log_b m$ to bring the exponent of $2$ down:

$$5 - 4 \log_{2(3x+5)} (x + 3)$$

### The Elegant Substitution

We define $a = \log_{(x+3)} [2(3x + 5)]$. By the base-change property, $\log_{2(3x+5)} (x + 3) = \frac{1}{a}$.

The equation transforms into:

$$a + 1 = 5 - \frac{4}{a}$$

Multiplying by $a$, we obtain $a^2 + a = 5a - 4$, which rearranges to:

$$a^2 - 4a + 4 = 0$$

This factors to $(a - 2)^2 = 0$, yielding the unique solution $a = 2$.

### Final Calculation and Rigor

Substituting back, we have $\log_{(x+3)} [2(3x + 5)] = 2$, which implies:

$$(x + 3)^2 = 2(3x + 5)$$

Expanding this, we get $x^2 + 6x + 9 = 6x + 10$. The $6x$ terms cancel out, leaving:

$$x^2 = 1 \Rightarrow x = 1, x = -1$$

We must verify the domain constraints for logarithms (base $> 0$ and $
eq 1$). Both $x = 1$ and $x = -1$ are valid.

The sum of the solutions is $1 + (-1) = \mathbf{0}$.

Animated Solution for Mathematics - Basic Mathematics: The sum of all the real solutions of the equation $lo g_{(x + 3)} (6 x^{2} + 28 x + 30) = 5 - 2 lo g_{(6 x + 10)} (x^{2} + 6 x + 9)$ is equal to

Select Answer:

Visualized Solution

Analyze the Equation

Factorize $6 x^{2} + 28 x + 30$

Factorize $x^{2} + 6 x + 9$

Factorize $6 x + 10$

Rewrite the Equation

Simplify LHS using $lo g (mn)$

Simplify RHS using $lo g (m^{n})$

Define Substitution $a$

Form the Equation in $a$

Solve for $a$

Solve for $x$

Check Domain Constraints

Final Sum of Solutions

The Sigma Insight: Logarithmic Equations and Inequalities

Analyzing the Setup

The Detective Work of Factorization

The Logarithmic Dance

The Elegant Substitution

Final Calculation and Rigor

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Animated Solution for Mathematics - Basic Mathematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of all the real solutions of the equation log(x+3)​(6x2+28x+30)=5−2log(6x+10)​(x2+6x+9) is equal to

Select Answer:

Visualized Solution

Analyze the Equation

Factorize .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6x2+28x+30

Factorize .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+6x+9

Factorize .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6x+10

Rewrite the Equation

Simplify LHS using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } log(mn)

Simplify RHS using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } log(mn)

Define Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Form the Equation in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x