JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 15)
The number of solutions of the equation
$${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$$, x > 0, is :
$${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$$, x > 0, is :
Answer
1
Explanation
$${\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} - 4 = 0$$
$${\log _{(x + 1)}}(2x + 5)(x + 1) + 2{\log _{(2x + 5)}}(x + 1) = 4$$
$${\log _{(x + 1)}}(2x + 5) + 1 + 2{\log _{(2x + 5)}}(x + 1) = 4$$
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
$${\log _{(x + 1)}}(2x + 5)(x + 1) + 2{\log _{(2x + 5)}}(x + 1) = 4$$
$${\log _{(x + 1)}}(2x + 5) + 1 + 2{\log _{(2x + 5)}}(x + 1) = 4$$
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
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