Question

Find the value of x satisfying the equation \[{\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} = 4\] .
A. -2
B. 2
C. -4
D. 4

Answer 1

Hints First solve the quadratic equation \[2{x^2} + 7x + 5 = 0\] then apply the formulas of logarithm in the given equation for simplification. Substitute t for \[{\log _{x + 1}}(2x + 5)\]in the obtained simplified equation and obtain a quadratic equation of t. Then solve the equation of t and ignore the negative values to obtain the required equation.

Formula used
Some formulas of logarithm are,
1)\[{\log _a}b = \dfrac{1}{{{{\log }_b}a}}\]
2)\[\log M + \log N = \log MN\]
3)\[{\log _a}a = 1\]
4)\[\log {a^m} = m\log a\]

Complete step by step solution
The given equation is,
\[{\log _{(x + 1)}}(2{x^2} + 7x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} = 4\]
\[{\log _{(x + 1)}}(2{x^2} + 5x + 2x + 5) + {\log _{(2x + 5)}}{(x + 1)^2} = 4\]
\[{\log _{(x + 1)}}\left[ {x(2x + 5) + 1(2x + 5)} \right] + 2{\log _{(2x + 5)}}(x + 1) = 4\]
\[{\log _{(x + 1)}}(x + 1)(2x + 5) + 2{\log _{(2x + 5)}}(x + 1) = 4\]
\[{\log _{(x + 1)}}(x + 1) + {\log _{(x + 1)}}(2x + 5) + 2{\log _{(2x + 5)}}(x + 1) = 4\]
\[1 + {\log _{(x + 1)}}(2x + 5) + 2.\dfrac{1}{{{{\log }_{(x + 1)}}(2x + 5)}} = 4\]
Substitute t for \[{\log _{(x + 1)}}(2x + 5)\] in the equation \[1 + {\log _{(x + 1)}}(2x + 5) + 2.\dfrac{1}{{{{\log }_{(x + 1)}}(2x + 5)}} = 4\]for further calculation.
\[1 + t + \dfrac{2}{t} = 4\]
\[t + {t^2} + 2 = 4t\]
\[{t^2} - 3t + 2 = 0\]
\[{t^2} - 2t - t + 2 = 0\]
\[t(t - 2) - 1(t - 2) = 0\]
\[(t - 1)(t - 2) = 0\]
\[t = 1,2\]
Now, calculate for \[t = 1\] .
\[{\log _{(x + 1)}}(2x + 5) = 1\]
\[2x + 5 = x + 1\]
\[x = - 4\]
Neglect the negative values of x as the function is a logarithm function.
Now, calculate for \[t = 2\] .
\[{\log _{(x + 1)}}(2x + 5) = 2\]
\[2x + 5 = {\left( {x + 1} \right)^2}\]
\[2x + 5 = {x^2} + 2\cdot x\cdot 1 + 1\]
\[{x^2} = 4\]
\[x = \pm 2\]
Neglect the negative values of x as the function is a logarithm function.
Therefore, \[x = 2\] .
The correct option is B.

Note If we take the negative values of x then the base of the logarithm becomes negative, which is not possible, so we are neglecting the negative values of x.