JEE Main 2026 — Basic of Mathematics Question with Solution

For the logarithms to be defined, we must satisfy the following conditions:
1. Base of the first logarithm: $x+1>0\Rightarrow x>-1$ and $x+1\ne 1\Rightarrow x\ne 0$.
2. Base of the second logarithm: $2x+3>0\Rightarrow x>-\frac{3}{2}$ and $2x+3\ne 1\Rightarrow x\ne -1$.
3. Arguments must be positive: $2x^2+5x+3>0$ and $x^2+2x+1>0$.
Taking the intersection of all these conditions, the domain of the equation is $x\in (-1,0)\cup (0,\infty )$.

Now, factorizing the arguments of the logarithms:
$2x^2+5x+3=(2x+3)(x+1)$
$x^2+2x+1=(x+1{)}^2$

Substitute these into the given equation:
${\log }_{(x+1)}((2x+3)(x+1))=4-{\log }_{(2x+3)}((x+1{)}^2)$

Using the properties of logarithms:
${\log }_{(x+1)}(2x+3)+{\log }_{(x+1)}(x+1)=4-2{\log }_{(2x+3)}(x+1)$
${\log }_{(x+1)}(2x+3)+1=4-\frac{2}{{\log }_{(x+1)}(2x+3)}$

Let $t={\log }_{(x+1)}(2x+3)$. The equation becomes:
$t+1=4-\frac{2}{t}$
$t-3+\frac{2}{t}=0$
$t^2-3t+2=0$
$(t-1)(t-2)=0$
$\Rightarrow t=1$ or $t=2$

Case 1: $t=1$
${\log }_{(x+1)}(2x+3)=1$
$2x+3=x+1$
$x=-2$
This value is rejected because $x=-2$ does not fall in the domain $x>-1$.

Case 2: $t=2$
${\log }_{(x+1)}(2x+3)=2$
$2x+3=(x+1{)}^2$
$2x+3=x^2+2x+1$
$x^2=2$
$x=\pm \sqrt{2}$

Since $x>-1$, $x=-\sqrt{2}$ is rejected. The only valid solution is $x=\sqrt{2}$.

The sum of squares of all the real solutions is $(\sqrt{2}{)}^2=2$.

Answer: $2$