Basic Maths and Logarithm
Logarithmic Equations
JEE Main 2021
Grade 11
Question:
The number of solutions of $\log_{x+1}(2x^2 + 7x + 5) + \log_{2x+5}(x + 1)^2 - 4 = 0$, $x > 0$, is:
(1) $0$
(2) $1$
(3) $2$
(4) $3$
Step-by-Step Solution
Key Concept: Factor: $2x^2 + 7x + 5 = (2x + 5)(x + 1)$. So $\log_{x+1}(2x + 5) + 1 + 2 \log_{2x+5}(x + 1) - 4 = 0$. Let $t = \log_{x+1}(2x + 5)$; then $t + 2/t = 3 \Rightarrow t = 1$ or $t = 2$. $t = 1 \Rightarrow x = -4$ (rejected). $t = 2 \Rightarrow (x + 1)^2 = 2x + 5 \Rightarrow x = 2$. One solution.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)