JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 21

Rewrite. With $t=\log_{x+1}(2x+3)$, the left side is $1+t$ and $\log_{2x+3}(x+1)^2=\tfrac2t$:
$$1+t=4-\frac2t\ \Rightarrow\ t^2-3t+2=0,\quad t=1\text{ or }2.$$

Cases.

• $t=1$: $2x+3=x+1$, so $x=-2$, which makes the base negative.
• $t=2$: $2x+3=(x+1)^2$, so $x^2=2$. Only $x=\sqrt2$ gives valid bases.

Sum of squares: 2.

Answer: 2