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Logarithms

Section: Mathematics

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Question : 9 of 11

Marks: +1, -0

The number of solutions of the equation

\log_{x+1}(2x^{2}+7x+5)+\log_{2x+5}(x+1)^{2}

-4=0

x>0

[20 Jul 2021 Shift 2]

Your Answer:

Solution:

\log_{x+1}(2x^{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+\frac{2}{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

\& \qquad 2x+5=(x+1)^{2}

x=-4

x^{2}=4 \Rightarrow x=2,-2

(rejected)

x=2

(rejected)

So,

=1

No. of solution

=1

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