JEE Main 2026 — Mathematics Algebra

For the logarithms to be defined, we must satisfy the following conditions:
1. Base of the first logarithm: $x+1 > 0 \Rightarrow x > -1$x+1>0⇒x>−1 and $x+1 \neq 1 \Rightarrow x \neq 0$x+1=1⇒x=0.
2. Base of the second logarithm: $2x+3 > 0 \Rightarrow x > -\dfrac{3}{2}$2x+3>0⇒x>−23​ and $2x+3 \neq 1 \Rightarrow x \neq -1$2x+3=1⇒x=−1.
3. Arguments must be positive: $2x^2+5x+3 > 0$2x2+5x+3>0 and $x^2+2x+1 > 0$x2+2x+1>0.
Taking the intersection of all these conditions, the domain of the equation is $x \in (-1, 0) \cup (0, \infty)$x∈(−1,0)∪(0,∞).
Now, factorizing the arguments of the logarithms:
$2x^2+5x+3 = (2x+3)(x+1)$2x2+5x+3=(2x+3)(x+1)
$x^2+2x+1 = (x+1)^2$x2+2x+1=(x+1)2
Substitute these into the given equation:
$\log_{(x+1)}((2x+3)(x+1)) = 4 - \log_{(2x+3)}((x+1)^2)$log(x+1)​((2x+3)(x+1))=4−log(2x+3)​((x+1)2)
Using the properties of logarithms:
$\log_{(x+1)}(2x+3) + \log_{(x+1)}(x+1) = 4 - 2\log_{(2x+3)}(x+1)$log(x+1)​(2x+3)+log(x+1)​(x+1)=4−2log(2x+3)​(x+1)
$\log_{(x+1)}(2x+3) + 1 = 4 - \dfrac{2}{\log_{(x+1)}(2x+3)}$log(x+1)​(2x+3)+1=4−log(x+1)​(2x+3)2​
Let $t = \log_{(x+1)}(2x+3)$t=log(x+1)​(2x+3). The equation becomes:
$t + 1 = 4 - \dfrac{2}{t}$t+1=4−t2​
$t - 3 + \dfrac{2}{t} = 0$t−3+t2​=0
$t^2 - 3t + 2 = 0$t2−3t+2=0
$(t-1)(t-2) = 0$(t−1)(t−2)=0
$\Rightarrow t = 1$⇒t=1 or $t = 2$t=2
Case 1: $t = 1$t=1
$\log_{(x+1)}(2x+3) = 1$log(x+1)​(2x+3)=1
$2x+3 = x+1$2x+3=x+1
$x = -2$x=−2
This value is rejected because $x = -2$x=−2 does not fall in the domain $x > -1$x>−1.
Case 2: $t = 2$t=2
$\log_{(x+1)}(2x+3) = 2$log(x+1)​(2x+3)=2
$2x+3 = (x+1)^2$2x+3=(x+1)2
$2x+3 = x^2+2x+1$2x+3=x2+2x+1
$x^2 = 2$x2=2
$x = \pm\sqrt{2}$x=±2​
Since $x > -1$x>−1, $x = -\sqrt{2}$x=−2​ is rejected. The only valid solution is $x = \sqrt{2}$x=2​.
The sum of squares of all the real solutions is $(\sqrt{2})^2 = 2$(2​)2=2.
Answer: $2$2