Question

The value of $x$ satisfying the equation ${\log }_{(x+1)}(2x^2+7x+5)+{\log }_{(2x+5)}{(x+1)}^2=4$is:

• A. $-2$
• B. $2$
• C. $-4$
• D. $4$

Answer 1

The correct option is B

$2$

Explanation for the correct option:

Find the required value.

Given: ${\log }_{(x+1)}(2x^2+7x+5)+{\log }_{(2x+5)}{(x+1)}^2=4$

So,

$$\begin{gathered} {\log }_{(x+1)}(2x^2+7x+5)+{\log }_{(2x+5)}{(x+1)}^2=4 \\ \Rightarrow {\log }_{(x+1)}(2x^2+2x+5x+5)+{\log }_{(2x+5)}{(x+1)}^2=4 \\ \Rightarrow {\log }_{(x+1)}\left[2x\left(x+1\right)+5(x+1)\right]+{\log }_{(2x+5)}{(x+1)}^2=4 \\ \Rightarrow {\log }_{(x+1)}\left[\left(2x+5\right)(x+1)\right]+{\log }_{(2x+5)}{(x+1)}^2=4 \\ \Rightarrow {\log }_{(x+1)}\left(2x+5\right)+{\log }_{(x+1)}(x+1)+{\log }_{(2x+5)}{(x+1)}^2=4\left[\because \log \left(mn\right)=\log m+\log n\right] \\ \Rightarrow {\log }_{(x+1)}\left(2x+5\right)+1+{\log }_{(2x+5)}{(x+1)}^2=4\left[\because {\log }_{a}a=1\right] \\ \Rightarrow {\log }_{(x+1)}\left(2x+5\right)+1+2{\log }_{(2x+5)}(x+1)=4\left[\because \log m^n=n\log m\right] \\ \Rightarrow {\log }_{(x+1)}\left(2x+5\right)+\frac{2}{{\log }_{(x+1)}\left(2x+5\right)}-3=0\left[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}\right] \end{gathered}$$

Put, ${\log }_{(x+1)}(2x+5)=t$

$$\begin{gathered} \Rightarrow t+\frac{2}{t}-3=0 \\ \Rightarrow \frac{t^2+2-3t}{t}=0 \\ \Rightarrow t^2+2-3t=0 \\ \Rightarrow t^2-2t-t+2=0 \\ \Rightarrow t\left(t-2\right)-1\left(t-2\right)=0 \\ \Rightarrow \left(t-1\right)\left(t-2\right)=0 \\ \Rightarrow t=1,2 \end{gathered}$$

For $t=1$

$$\begin{gathered} {\log }_{(x+1)}(2x+5)=1 \\ \Rightarrow (2x+5)={(x+1)}^1\left[\because {\log }_{a}b=n\Rightarrow b=a^n\right] \\ \Rightarrow x=-4 \end{gathered}$$

Which gives the negative base of ${\log }_{(-4+1)}={\log }_{-3}$ so $x=-4$is not possible

Now for $t=2$

$$\begin{gathered} {\log }_{(x+1)}(2x+5)=2 \\ \Rightarrow (2x+5)={(x+1)}^2\left[\because {\log }_{a}b=n\Rightarrow b=a^n\right] \\ \Rightarrow 2x+5=x^2+2x+1 \\ \Rightarrow 2x+5=x^2+2x+1 \\ \Rightarrow x^2=4 \\ \Rightarrow x=+2,x=-2 \end{gathered}$$

$x=-2$ gives the negative base of ${\log }_{(-2+1)}={\log }_{-1}$ so $x=-2$is not possible

Therefore the required value of $x$ is $2$

Hence, option (B) is the correct answer.