The number of solutions of the equation log(x+1)(2x² + 7x + 5) + log(2x+5)(x + 1)² - 4 = 0, x 0, i

Q: Wait, why do we reject the negative roots $x = -4$ and $x = -2$?

Great question! In logarithmic equations, the base must be positive and not equal to 1, and the argument must be strictly positive. The problem explicitly states $x > 0$. Even if a value satisfies the algebraic equation, if it violates the domain constraints (like making the base or argument non-positive), it must be rejected. Always check your final answers against the original domain!

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

This is a pure mathematics problem, not physics! While physics problems often use energy conservation, here we rely on the fundamental properties of logarithms. The 'shortcut' is recognizing the reciprocal relationship between the two logarithmic terms, which allows us to transform a complex transcendental equation into a simple quadratic one.

Q: Is there a specific reason we chose $t = \log_{(x+1)}(2x + 5)$?

Yes! By observing the structure of the equation, we saw that the two logarithmic terms were inverses of each other. Choosing $t$ as one of them turns the equation into a standard quadratic form $t^2 - 3t + 2 = 0$, which is much easier to solve than working with logs directly. It is a classic substitution strategy used to simplify complex expressions.

Question

The number of solutions of the equation $\log_{(x+1)}(2x^2 + 7x + 5) + \log_{(2x+5)}(x + 1)^2 - 4 = 0$, $x > 0$, is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of the mathematical landscape! Today, we are standing before a seemingly formidable logarithmic equation:

$$\log_{(x+1)}(2x^2 + 7x + 5) + \log_{(2x+5)}(x + 1)^2 - 4 = 0$$

It looks like a tangled mess of variables, bases, and exponents, but in the world of JEE Advanced, complexity is often just a mask for hidden elegance. Our mission is to peel back that mask.

First, we must acknowledge the silent guardian of this problem: the domain constraints. For the logarithms to be defined, the bases must be positive and not equal to $1$, and the arguments must be positive. Specifically, $x+1 > 0$, $x+1 
eq 1$, $2x+5 > 0$, and $2x+5 
eq 1$.

### The Hidden Symmetry

Let us focus our gaze on the argument of the first logarithm: $2x^2 + 7x + 5$. If we apply the middle-term splitting method, we find that $2x^2 + 5x + 2x + 5$ factors beautifully into $(2x + 5)(x + 1)$.

Suddenly, the fog begins to lift. Look at the bases of our logarithms: $(x+1)$ and $(2x+5)$. They are the exact factors we just uncovered!

By substituting this back, our equation transforms into:

$$\log_{(x+1)}((2x+5)(x+1)) + \log_{(2x+5)}(x + 1)^2 - 4 = 0$$

### The Logarithmic Dance

Now, we invoke the powerful tools in our arsenal. Using the product rule, $\log_b(mn) = \log_b m + \log_b n$, we expand the first term into $\log_{(x+1)}(2x + 5) + \log_{(x+1)}(x + 1)$.

Since $\log_a a = 1$, the second part simplifies to $1$. Simultaneously, we use the power rule, $\log_b(a^k) = k \log_b a$, on the second term to bring the exponent $2$ down, giving us $2 \log_{(2x+5)}(x + 1)$.

Our equation now breathes much easier:

$$\log_{(x+1)}(2x + 5) + 1 + 2 \log_{(2x+5)}(x + 1) - 4 = 0$$

This simplifies to:

$$\log_{(x+1)}(2x + 5) + 2 \log_{(2x+5)}(x + 1) - 3 = 0$$

### The Masterstroke: Substitution

Observe the two logarithmic terms. One has base $(x+1)$ and argument $(2x+5)$, while the other is the exact reciprocal. This is the moment of clarity.

Let $t = \log_{(x+1)}(2x + 5)$. Then the second term is simply $\frac{2}{t}$. The equation becomes:

$$t + \frac{2}{t} - 3 = 0$$

Multiplying by $t$, we arrive at the quadratic:

$$t^2 - 3t + 2 = 0$$

Factoring this gives $(t - 1)(t - 2) = 0$, yielding $t = 1$ or $t = 2$.

### The Final Verdict

Now, we solve for $x$. If $t = 1$, then $\log_{(x+1)}(2x + 5) = 1$, implying $x + 1 = 2x + 5$, which leads to $x = -4$.

If $t = 2$, then $\log_{(x+1)}(2x + 5) = 2$, implying $(x + 1)^2 = 2x + 5$. This simplifies to $x^2 + 2x + 1 = 2x + 5$, or $x^2 = 4$, giving $x = 2$ or $x = -2$.

Finally, we return to our domain constraints. We must reject $x = -4$ and $x = -2$ because they violate the base requirements ($x+1 > 0$ and $x+1 
eq 1$).

Only **$x = 2$** remains standing. There is exactly **$1$** valid solution. You have conquered the jungle!

Animated Solution for Mathematics - Basic Mathematics: The number of solutions of the equation $lo g_{(x + 1)} (2 x^{2} + 7 x + 5) + lo g_{(2 x + 5)} (x + 1)^{2} - 4 = 0$ , $x > 0$ , is

Enter Numerical Value:

Visualized Solution

Analyzing the Logarithmic Equation

Factorizing the Quadratic Argument

Rewriting the Equation

Applying the Logarithm Product Rule

Simplifying the Same Base Logarithm

Applying the Logarithm Power Rule

Combining Constants

The Substitution Strategy

Forming the Quadratic Equation in t

Solving for t

Case 1: Evaluating t = 1

Case 2: Evaluating t = 2

Checking Domain Constraints and Final Answer

The Sigma Insight: Logarithmic Equations and Inequalities

Analyzing the Setup

The Hidden Symmetry

The Logarithmic Dance

The Masterstroke

The Final Verdict

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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 number of solutions of the equation log(x+1)​(2x2+7x+5)+log(2x+5)​(x+1)2−4=0, x>0, is