What are the real solutions of log_(x+1)(2x²+5x+3)=4−log_(2x+3)(x²+2x+1)?

Only x=√2. x=−2 and x=−√2 are rejected due to invalid logarithm bases.

What is the sum of squares of real solutions?

(√2)²=2. There is only one valid solution x=√2.

Sum of squares of all real solutions of log_(x+1)(2x²+5x+3)=4−log_(2x+3)(x²+2x+1)

JEERANKERS

MathLogarithms

QInteger TypeLogarithms

The sum of squares of all the real solutions of the equation $$\log_{(x+1)}(2x^2+5x+3)=4-\log_{(2x+3)}(x^2+2x+1)$$ is equal to __________.

✅ Correct Answer

✓

Solution

Factorise the arguments

$2x^2+5x+3=(2x+3)(x+1)$
$x^2+2x+1=(x+1)^2$

Domain: $x+1>0,\neq1$ and $2x+3>0,\neq1$. So $x>0$ (from $x>-1,x\neq0$ and $x>-1,x\neq-1$).

Simplify using log properties

$\log_{x+1}[(2x+3)(x+1)]=4-\log_{2x+3}(x+1)^2$

$\log_{x+1}(2x+3)+\underbrace{\log_{x+1}(x+1)}_{=1}=4-2\log_{2x+3}(x+1)$

$\log_{x+1}(2x+3)+1=4-2\log_{2x+3}(x+1)$

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

Substitute t = log_(x+1)(2x+3)

Note: $\log_{2x+3}(x+1)=1/t$.

$t-3=-\dfrac{2}{t}$

$t(t-3)=-2$

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

$(t-1)(t-2)=0\quad\Rightarrow\quad t=1\text{ or }t=2$

Case t = 1: log_(x+1)(2x+3) = 1

$2x+3=x+1\quad\Rightarrow\quad x=-2$

Check domain: $x=-2$ gives base $x+1=-1<0$. Rejected.

Case t = 2: log_(x+1)(2x+3) = 2

$2x+3=(x+1)^2=x^2+2x+1$

$x^2-2=0\quad\Rightarrow\quad x=\pm\sqrt{2}$

Check $x=\sqrt{2}$: base $x+1=1+\sqrt{2}>1$ ✓, $2x+3>0$ ✓. Valid.

Check $x=-\sqrt{2}$: base $x+1=1-\sqrt{2}<0$. Rejected.

Sum of squares of valid solutions

$$\text{Only valid solution: }x=\sqrt{2}\quad\Rightarrow\quad(\sqrt{2})^2=\boxed{2}$$

📘 Logarithmic Equation Strategy

When both bases involve $x$, let $t=\log_a b$ and use $\log_b a=1/t$. This converts the equation to a polynomial in $t$. Always verify domain conditions: base must be positive and $\neq1$, argument must be positive.

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Important Concepts

Factorisation is Key$2x^2+5x+3=(2x+3)(x+1)$ and $x^2+2x+1=(x+1)^2$. Splitting the log of a product using $\log(AB)=\log A+\log B$ reduces the equation significantly.

Change of Base Reciprocal$\log_{x+1}(2x+3)=t\Rightarrow\log_{2x+3}(x+1)=1/t$. This is valid as long as both bases are positive and $\neq1$, which is handled by the domain check.

Domain Must Be CheckedBase $x+1>0,\neq1$ requires $x>-1,x\neq0$. Also $2x+3>0,\neq1$ requires $x>-3/2,x\neq-1$. Combined with arguments positive: $x>0$ is the safe domain.

Rejecting Extraneous Solutions$x=-2$ fails because $x+1=-1<0$ (invalid base). $x=-\sqrt{2}$ fails because $x+1=1-\sqrt{2}<0$. Only $x=\sqrt{2}$ survives all checks.

FAQs

Why substitute t = log_(x+1)(2x+3)?

⌄

It relates the two different log bases: $\log_{x+1}(2x+3)=t$ and $\log_{2x+3}(x+1)=1/t$ (reciprocal property). This converts the equation to a simple quadratic $t^2-3t+2=0$.

How to get t−3=−2/t?

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From step 2: $t+1=4-2/t$ becomes $t+1-4=-2/t$, i.e., $t-3=-2/t$. Multiply both sides by $t$: $t^2-3t=-2$, giving $t^2-3t+2=0$.

Why is x=−√2 rejected?

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Base $=x+1=1-\sqrt{2}\approx1-1.414=-0.414<0$. A logarithm base must be positive and $\neq1$. Negative base is not allowed.

Can the sum of squares be 0?

⌄

Only if $x=0$ were the solution. But $x=0$ is outside the domain (base $x+1=1$ is not allowed). The only solution is $x=\sqrt{2}$, giving sum $=2$.

Is this from JEE Main 2026?

⌄

Yes, this question appeared in JEE Main 2026 Section B.

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