${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$ ${\log _{(x + 1)}}(2x + 5)(x + 1) + 2{\log _{(2x + 5)}}(x + 1) = 4$ ${\log _{(x + 1)}}(2x + 5) + 1 + 2{\log _{(2x + 5)}}(x + 1) = 4$ Put ${\log _{(x + 1)}}(2x + 5) = t$ $t + {2 \over t} = 3 \Rightarrow {t^2} - 3t + 2 = 0$ t = 1, 2 ${\log _{(x + 1)}}(2x + 5) = 1$ & ${\log _{(x + 1)}}(2x + 5) = 2$ $x + 1 = 2x + 3$ & $2x + 5 = {(x + 1)^2}$ $x = - 4$ (rejected) ${x^2} = 4 \Rightarrow x = 2, - 2$ (rejected) So, x = 2 No. of solution = 1